
Class QP\^05- 
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Book kli> 



ELEMENTS 



OF THE 



DIFFERENTIAL AND INTEGRAL 

CALCULUS. 

BY ELIAS LOO MIS, LLD., 

PROFESSOR OF NATURAL PHILOSOPHY AND ASTRONOMY IN YALE COLLEGE, AND AUTHOR OF 
"A COURSE OF MATHEMATICS." 



REVISED EDITION. 




NEW YORK: 
HARPER & BROTHERS, PUBLISHERS, 

FRANKLIN SQUARE. 

18 74. 



OlK 



n^ 






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PREFACE. 



In preparing the first edition of my Differential and In- 
tegral Calculus, published in 1851, my object was to exhibit 

j elementary principles of this branch of science in a form 
j simple that they might be distinctly apprehended by begin- 

rs who had no special aptitude for mathematical studies. 
1\at this attempt was in some measure successful is believed 

be indicated by the number of copies of the book (25,000) 
which have been sold up to 1874. It is, however, well known 
that within the last few years not only has the number of sci- 
entific schools in our country greatly increased, but the courses 
of instruction have been rendered more thorough and com- 
plete. Under these circumstances, I have felt that the former 
edition of my book was too restricted in its plan, and I de- 
cided to rewrite it, so as to accommodate it to the present con- 
dition of our scientific schools. In doing this, I have labored 
to develop the fundamental principles in a method which is 
strictly logical, and at the same time easily apprehended by 
students of ordinary capacity. I have also aimed to introduce 
as great a variety of topics as can be advantageously studied 
in the time generally devoted to it in our best colleges, and 
have selected those principles which are best adapted to show 
the utility of this branch of the Mathematics. For this pur- 
pose I have ventured somewhat beyond the limits ordinarily 
assigned to treatises on the Calculus, and have shown the ap- 
plication of these methods to a few practical questions in Me- 



IV PREFACE. 

chanics. Throughout the entire work I have aimed to illus- 
trate every principle by appropriate examples designed to test 
the clearness of the student's conceptions. It is hoped that in 
its present form the book may be found useful, not only to 
those who study the Calculus simply as a means of mental dis- 
cipline, but also to those who may have occasion to apply its 
principles to the solution of various problems in Mechanics 
and Astronomy; and that those who wish to prosecute the 
subject more extensively may find this volume a suitable in- 
troduction to more complete treatises on the Calculus. 

I take pleasure in acknowledging my special obligations to 
Professor H. A. Newton for the interest which he has mani- 
fested in the revision of this book. He has not only read the 
entire manuscript, but has also read all the proofs, and has 
made a very large number of suggestions of which I have stu- 
diously availed myself. If the present edition should be found 
superior to those which have preceded it, the difference must 
be ascribed in no small degree to the assistance which has 
thus been rendered me. 



CONTENTS. 



Page 

Historical Sketch 9 

DIFFERENTIAL CALCULUS. 

CHAPTER I. 
DEFINITIONS. LIMITING RATIOS. DIFFERENTIATION OF ALGEBRAIC 

FUNCTIONS. 

Definitions — Variables and Constants , 17 

Functions — Algebraic and Transcendental ; Increasing and Decreasing 19 

Limiting Values of Quantities 20 

Increments and their Limiting Ratio 23 

Differential Defined ; Differential Coefficient 25 

General Method of Differentiation 27 

Differential of any Power of a Variable 28 

Differential of the Sum or Difference of Several Functions 30 

Differential of the Product of Several Functions 34 

Differential of a Fraction 35 

Differential of a Function with any Exponent 36 

Differential of a Polynomial raised to any Power 39 

CHAPTER II. 
DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 

Differential of a Logarithm 46 

Illustration from Logarithmic Tables 49 

Differential of an Exponential Function 50 

Differential of a Trigonometrical Function 52 

Geometrical Illustration 57 

Differentials of Logarithmic Sine, Cosine, etc 58 

Inverse Trigonometrical Functions 60 

CHAPTER III. 

SUCCESSIVE DIFFERENTIATION. EXPANSION OF FUNCTIONS IN SERIES. 

FUNCTIONS OF SEVERAL INDEPENDENT VARIABLES. 

VANISHING FRACTIONS. 

Second Differential Coefficient 64 

Maclaurin's Theorem — Applications 67 

When Maclaurin's Theorem Fails 72 

Partial Differential Coefficients 73 

Taylor's Theorem — Applications 74 



VI CONTENTS. 

Page 

Differentiation of Functions of Two Independent Variables 79 

Total Differential of a Function 81 

Differential Equations 83 

Vanishing Fractions 85 

To Find the Value of a Vanishing Fraction 88 

CHAPTER IV. 
GEOMETRICAL REPRESENTATION OF THE FIRST DIFFERENTIAL COEF- 
FICIENT. MAXIMA AND MINIMA VALUES OF A FUNCTION. 

Signification of the First Differential Coefficient 91 

Maxima and Minima of Functions of a Single Variable .* 94 

Method of Finding Maxima and Minima 95 

Sign of the Second Differential Coefficient 99 

General Rule for Maxima and Minima 1 00 

How the Process of Finding Maxima and Minima may be Abbreviated 103 

Geometrical Problems in Maxima and Minima 106 

CHAPTER V. 
TANGENTS AND NORMALS TO PLANE CURVES. ASYMPTOTES. 

Equation of a Tangent Line 113 

Length of Tangent, Subtangent, Normal, and Subnormal 115 

Formulas Applied to Particular Curves 116 

Asympto f es to Curves — how Determined 119 

Tangents to Curves Referred to Polar Co-ordinates 125 

CHAPTER VI. 
CONVEXITY AND CONCAVITY. SINGULAR POINTS OF CURVES. TRACING 

OF CURVES. 

When a Curve is Convex toward the Axis of Abscissas 129 

When a Curve is Concave toward the Axis of Abscissas 130 

Singular Points of Curves * 132 

To Determine Points of Inflection 133 

To Determine Multiple Points 135 

To Determine whether a Curve has Cusps 137 

To Determine a Conjugate Point 140 

To Trace a Plane Curve from its Equation 142 

CHAPTER VII. 

DIFFERENTIAL COEFFICIENT OF AN ARC, AREA, ETC. 

Limiting Ratio of an Arc to its Chord 150 

Differential of an Arc of a Plane Curve ; of a Plane Area 151 

Differential of the Surface and Volume of a Solid of Revolution 168 

Differential of an Arc of a Polar Curve ; Area of a Polar Curve 156 

CHAPTER VIII. 

CONTACT. CURVATURE. EVOLUTES AND INVOLUTES. THE CYCLOID. 

I different Orders of ( lontact 159 

Osculating ( ircle Determined 102 



CONTENTS. Vll 

Page 

Radius of Curvature, how Found 164 

Evolutes and Involutes 166 

E volute of the Common Parabola, Circle, etc 167 

The Cycloid ; its Equation and Properties 172 

Equation of the Evolute of a Cycloid 1 74 



INTEGEAL CALCULUS. 

CHAPTER I. 

INTEGRATION OF MONOMIAL DIFFERENTIALS. OF CERTAIN BINOMIAL DIF- 
FERENTIALS. DEFINITE INTEGRALS. TRIGONOMETRICAL FUNCTIONS. 

Integral Calculus Defined 177 

Integral of the Sum or Difference of Several Differentials , 179 

Integration of Monomial Differentials 1 80 

Particular Binomial Differentials 183 

Definite Integral ; Integral between Limits 186 

Differentials of Trigonometrical Functions 189 

Differential of Arc in Terms of Sine, Cosine, Tangent, etc 1 9 L 

CHAPTER II. 
RATIONAL FRACTIONS. INTEGRATION OF CERTAIN IRRATIONAL DIFFER- 
ENTIALS. 

How a Rational Fraction can be Integrated 197 

Case in which the Denominator can be Resolved into Real Factors 197 

Case in which the Denominator contains Imaginary Factors 202 

Reduction of Irrational Differentials to Rational 205 

When the Irrational Parts are Trinomials • 208 

CHAPTER III. 

INTEGRATION OF BINOMIAL DIFFERENTIALS. INTEGRATION BY PARTS. 

FORMULAS OF REDUCTION. 

When a Binomial Differential can be Integrated 211 

Integration by Parts 217 

Methods of Reduction 220 

Formulas for the Reduction of Binomial Differentials 222 

Examples of their Application 225 

CHAPTER IV. 

INTEGRATION BY INFINITE SERIES. INTEGRATION OF LOGARITHMIC, 

EXPONENTIAL, AND TRIGONOMETRICAL DIFFERENTIALS. 

Integration by Infinite Series 229 

Integration of Logarithmic Differentials 230 

Integration of Exponential Differentials 233 

Integration of Trigonometrical Differentials 237 

Integration of Inverse Trigonometrical Differentials 211 



Vlll CONTENTS. 



CHAPTER V. 
SUCCESSIVE INTEGRATION. INTEGRATION OF FUNCTIONS OF TWO OR 

MORE VARIABLES. DIFFERENTIAL EQUATIONS. Page 

Successive Integration 243 

Integration of Differential Functions of Two Variables 244 

Differential Equations, how Classified 246 

How Differential Equations may be Solved 248 

Geometrical Applications of Differential Equations 251 

CHAPTER VI. 

QUADRATURE OF PLANE SURFACES. RECTIFICATION OF PLANE CURVES. 

SURFACES AND VOLUMES OF SOLIDS OF REVOLUTION. 

Quadrature of Plane Surfaces Referred to Rectangular Co-ordinates 253 

Quadrature of Polar Curves 259 

Area included between Two Plane Curves 26L 

Rectification of Plane Curves Referred to Rectangular Co-ordinates 264 

Rectification of Plane Curves Referred to Polar Co-ordinates 269 

Quadrature of Surfaces of Revolution 271 

Cubature of Solids of Revolution 276 

CHAPTER VII. 
APPLICATIONS OF THE DIFFERENTIAL AND INTEGRAL CALCULUS TO 

MECHANICS. 

Centre of Gravity of a Line 281 

Centre of Gravity of a Plane Surface 284 

Centre of Gravity of a Surface of Revolution and a Solid of Revolution 286 

Centre of Oscillation of a Pendulum, how Found 290 

( )n the Attraction of Bodies . 295 

To Find the Attraction of a Homogeneous Sphere 299 

Differentia] Equations for Force and Motion 301 

Fall of Bodies to the Earth— Time and Velocity Computed 302 

Attraction of a Spherical Shell on a Particle situated Within it 306 

Time of a Falling Body beneath the Surface of the Earth 309 



SKETCH OF THE HISTORY 



THE DIFFERENTIAL AND INTEGRAL CALCULUS. 



The invention of the Differential and Integral Calculus dates no further 
back than the seventeenth century, although the questions which led to 
its discovery must have presented themselves to the earliest geometers. 
When the ancient geometers wished to compare figures bounded by 
curved lines with figures bounded by straight lines, they were obliged to 
resort to indirect and prolix modes of demonstration. Euclid proved 
that circles are to one another as the squares of their diameters, by prov- 
ing that the square of one diameter can not be to the square of another 
diameter, as the one circle to a space either less or greater than the other 
circle. To accomplish this, he employed inscribed and circumscribed 
polygons, and proved that if the number of sides of the polygons be con- 
tinually doubled, the two figures must approach nearer and nearer to 
equality, and consequently nearer to the intermediate circle. This proc- 
ess has been called the Method of Exhaustions. 

By methods somewhat similar, Archimedes solved problems still more 
difficult — such as the determination of the ratio of the sphere to its cir- 
cumscribing cylinder, the quadrature of the parabola, and the properties 
of spirals. 

For eighteen centuries from the time of Archimedes no important prog- 
ress was made in this department of science. The next great improve- 
ment was made by Cavalieri of Italy, who, in 1635, published a new 
method, called the Method of Indivisibles. Cavalieri considered lines as 
made up of an infinite number of small points ; surfaces as composed of 
an infinite number of parallel lines, the breadths of which are indefinitely 
small ; and solids as made up of an infinite number of parallel planes, or 
rather of indefinitely thin plates or laminae. From this view, it follows 
that the magnitude of all lines, surfaces, and solids must be proportional 
to the number of elementary particles they contain ; and the only diffi- 
culty in comparing magnitudes by this method consists in dividing them 
into these elements and finding the relative numbers into which they can 
be divided. The accuracy of this method was disputed by Guldinus in 
1640 ; but the correctness of its results could be demonstrated by the 
method of exhaustions, and it afforded in many cases a ready means of 
ascertaining the ratio of curvilinear areas, and of solids bounded by 
curved surfaces. 

A2 



10 SKETCH OF THE HISTORY OP 

About the same time, Roberval in France discovered a method of solv- 
ing problems relating to curvilinear figures, which did not differ materi- 
ally from that of Cavalieri. 

In 1637, Descartes published a simple method of drawing tangents to 
all algebraic curves ; and by applying Algebra to express the relations of 
variable quantity, he made an entire revolution in the science of Mathe- 
matics, and prepared the way for the grand discoveries which soon fol- 
lowed. 

In the Arithmetic of Infinites, published by Wallis in 1655, was given the 
first application of algebraic calculation to quadratures, and this was 
founded on the method of indivisibles. By this method, Wallis demon- 
strated the fundamental rule for the quadrature of curves, whose ordinate 
is proportional to any power of the abscissa. This enabled him to square 
any curve in which the value of one of the co-ordinates can be expressed 
in terms of the other, without involving either fractional or negative ex- 
ponents. 

About the year 1666, Newton invented the method of Fluxions, by 
which he was able not only to draw tangents to mechanical curves, but 
also to solve all problems relating to the quadrature or rectification of 
curves. According to Newton, all quantities are supposed to be generated 
by continuous motion : a line, by the motion of a point ; a surface, by the 
motion of a line either constant or variable ; and a solid, by the motion 
of a plane which either retains the same magnitude, or varies according 
to a certain law. All other quantities, in whatever way they may be ex- 
pressed, are conceived to be generated in a similar manner. The relative 
rates or velocities with which magnitudes, and the variables on which 
they depend, are increasing at any instant, are termed their fluxions ; and 
the whole quantities generated in consequence of such velocities of in- 
crease are termed the fluents. 

In 1669, Professor Barrow, of Cambridge, announced to the Secretary of 
the Royal Society, that Newton had put into his hands a manuscript 
treatise containing general methods of calculating the quadrature and 
rectification of curves, and also of resolving equations. In this man- 
uscript the method of fluxions was first indicated ; and, through Professor 
Barrow, these discoveries were communicated not only to English mathe- 
maticians, but to several mathematicians on the Continent. This treatise 
was not published till 1711 — more than forty years after it was written. 

The first work in which Newton communicated any thing to the public 
on the subject of fluxions, w T as in the first edition of the "Principia," in 1687. 
The principle of his new calculus was there pointed out, but nothing ap- 
peared to indicate the peculiar notation which is so essential to this cal- 
culus. This first became known to the world in 1693, from the second 
volume of Wallis's works; and it was not till 1704 that Newton himself 
published a tract explaining the principles of his method, and applying it 
to quadratures. 



THE DIFFERENTIAL AND INTEGRAL CALCULUS. 11 

About the same period, the celebrated Leibnitz invented the Differen- 
tial and Integral Calculus, which corresponds to the direct and inverse 
method of fluxions. In 1673, Leibnitz visited London and communicated 
to several members of the Royal Society certain researches in the theory 
of the differences of numbers, and he was informed that this subject had 
already been treated by Mouton, a geometer of Lyons. He then turned 
his attention to the subject of infinite series, and in 1674 he announced to 
Oldenburg that he had discovered the series which gives the value of a 
circular arc in terms of its tangent, and that he had very general analytic 
methods. Oldenburg, in reply, informed him that Newton had discovered 
methods which give the quadrature of curves, both geometrical and me- 
chanical, and which extended to the circle. 

In June, 1676, Newton sent a letter to Oldenburg to be transmitted to 
Leibnitz. This letter contained his celebrated binomial theorem, and a 
variety of other matters relating to infinite series and quadratures, but 
nothing directly relating to the theory of fluxions. In a second letter, 
written in October, 1676, he explains the manner in which he found the 
binomial theorem, and suggests the character of his method of fluxions, 
but conceals the method by transposing the letters of the following sen- 
tence : " An equation containing fluent* Icing given, to find their fluxions and 
conversely ■." This letter is positive evidence that in 1676 Xewton was not 
only in possession of the principle of his calculus, but that he had already 
adopted the same nomenclature to which he subsequently adhered. 

In June, 1677, Leibnitz sent to Oldenburg, to be communicated to Xew- 
ton, a letter giving an account of a method which applies to every thing 
that could be done by the fluxions of Newton. This method was his dif- 
ferential cedculus. This letter not merely explains the principles of his 
calculus, but employs the same notation which is in use at the present 
day. In 1684, Leibnitz first published his method in a tract containing 
only six pages. 

According to the views of Leibnitz, all quantities are supposed to be 
composed of indefinitely small portions admitting of all possible degrees 
of relative magnitude. Such portions as are indefinitely small compared 
to any finite magnitude, are termed infinitesimals of the first order ; such as 
are indefinitely small compared to these are called inflnitesinuds of the 
second order, and so on. Areas are accordingly supposed to be made up 
of parallelograms having one side finite and another indefinitely small ; 
solids are supposed to be made up of parallelopipeds of finite base but 
indefinitely small altitudes, etc. ; and they are compared with each other 
by finding the sums of such indefinitely small rectangles, parallelopipeds, 
etc., by methods which the calculus itself suggests. 

Some persons objected to this analysis, because it has the appearance 
of being only an approximation, and of not coming up to the perfect 
standard of geometrical precision. Newton endeavored to preserve his 
calculus froin the imputation of throwing any thing away merely because 



12 SKETCH OF THE HISTORY OF 

it was small. Instead of the actual increments of the variable quantities, 
he employed what he called the fluxions of those quantities ; meaning, by 
fluxions, quantities which had to one another the same ratio which the 
increments had in their ultimate state. He did not, therefore, reject 
quantities merely because they were so small that he might do so without 
committing any sensible error, but because he must reject them in order 
to commit no error whatsoever. Fluxions were with him nothing else 
than measures of the velocities with which variable quantities were sup- 
posed to be generated. 

When the principles of the differential calculus were attacked, Leibnitz 
compared his method with that of Archimedes, and showed that it was 
essentially the same ; for instead of supposing the differentials to be in 
fact infinitely small, it is only necessary to conceive that we may always 
take them so small that the error which results from the quantities omit- 
ted in the calculus shall be less than any given quantity. Indeed, the 
fluxions and fluents of Newton correspond essentially to the differentials 
and integrals of Leibnitz ; so that the two methods differ only in the no- 
tation, and in the peculiar modes of viewing the subject, or what is com- 
monly called the metaphysics of the calculus. 

The facts which have thus been briefly stated have generally been re- 
garded as proving that Newton was the first inventor of that analysis, 
which he called by the name of fluxions ; but that in the communications 
made by him or his friends to Leibnitz there was nothing that could con- 
vey any idea of the nature of that method, or of the principles on which 
that analysis was founded. Leibnitz is, therefore, regarded as the second 
discoverer of the new analysis, but to have made the discovery entirely 
independent of Newton, and to have an equal claim to originality. Leib- 
nitz has also the merit of having first made known his discovery to the 
world, having published an account of it in 1684. 

Leibnitz enjoyed without contradiction the honor of being the inventor 
of his calculus until 1G99, when a mathematician named Duillier, in a 
short tract on the curve of swiftest descent, asserted that Newton was the 
first inventor of the new calculus, and that he would leave to others to 
decide whether Leibnitz had borrowed any thing from the English geom- 
eter. To this charge Leibnitz replied, declaring himself, equally with 
Newton, to have been the inventor, and asserting his own claim to the 
first publication of the calculus. 

In 1704, when Newton's treatise on the quadrature of curves was pub- 
lished, the Leipsie journalists gaye an unfavorable account of it, and inti- 
mated that Newton had been led to the idea of fluxions by the differen- 
tials of Leibnitz. This charge roused the indignation of the British math- 
ematicians, and in 1708 Keill undertook to prove that the communica- 
tions of Newton to Leibnitz were sufficient to put the latter in possession 
of the principles of the new analysis, and that he had only substituted the 
notion of differentials lor that of fluxions. 



TIIE DIFFERENTIAL AND INTEGRAL CALCULUS. 13 

Leibnitz complained of KehTs proceedings to the Royal Society of Lon- 
don, and the society appointed a commission of its members to draw up 
a full report of all the communications which had passed between Newton 
and Leibnitz or their friends on subjects connected with the new analysis. 
In this report the committee avoided expressing any direct opinion upon 
the question whether Leibnitz had invented the calculus for himself, or 
had availed himself of the labors of Newton ; but the report seemed to in- 
dicate that they were of the latter opinion. Some of the friends of Leib- 
nitz endeavored to fasten upon Newton the charge of plagiarism, and de- 
clared that there was no reason to believe that the fluxionary calculus was 
invented before the differential. Now that the excitement occasioned by 
this controversy has passed away, mathematicians are unanimous in con- 
ceding to Newton the claim of first discoverer ; and they also concede 
that Leibnitz was an independent discoverer. They also concede to Leib- 
nitz the credit of priority in the publication of his new method, and of 
having invented a superior system of notation, which has entirely super- 
seded the notation introduced by Newton. 

In order to attract the attention of mathematicians to the importance 
of the new analysis, Leibnitz, in 1687, proposed the following problem : 
" To determine the curve a heavy body must describe in order to descend 
equally in equal times." This problem was resolved by Huygens and 
James Bernoulli. In 1690, Bernoulli proposed the problem of the curve 
which a chain of uniform weight makes when suspended from two 
points. This problem was resolved by Huygens, Leibnitz, and the pro- 
poser. In 1697, John Bernoulli proposed the problem concerning the 
line along which a body must descend in order to go from one point to 
another, not perpendicularly under it, in the least possible time. This 
problem was resolved by Leibnitz, Newton, the two Bernoullis, and De 
THopital. In 1716, Leibnitz proposed the problem, "A system of curves 
described according to a known law being given, to describe a curve 
which shall cut them all at right angles." This problem was resolved by 
Newton and John Bernoulli. In 1718, Keill, an English mathematician, 
challenged John Bernoulli to find the path of a projectile moving in a 
medium in which the resistance varies as the square of the velocity. Ber- 
noulli resolved the problem not only in that particular case, but also 
when the resistance varies as any power of the velocity. He then re- 
quired that Keill should produce his own solution ; but Keill found his 
problem too difficult, and maintained a profound silence. 

The first elementary treatise in which the principles of the new calcu- 
lus were explained was the " Analyse des Infiniment Petits," by the Mar- 
quis De THopital, published in 1696, from which time a knowledge of the 
new analysis became generally diffused. It was not, however, every where 
accepted without opposition. Some mathematicians of considerable em- 
inence in Germany, France, and England objected to the notion of quan- 
tities infinitely small. Among those who attacked the new analysis was 



14 SKETCH OF THE HISTORY OF 

Bishop Berkeley. He endeavored to prove that this analysis was inaccu- 
rate in its principles, and that if it ever leads to true conclusions, it is 
from an accidental compensation of errors ; and he attempted to fortify 
his objections by ridicule, calling ultimate ratios the ghosts of departed 
quantities. The objections of Berkeley were fully answered by Jurin and 
others. 

The succession of great geometers was not interrupted by the death of 
Newton, Leibnitz, and the Bernoullis. Euler and D'Alembert, Lagrange 
and Laplace, Gauss and Abel, Cauchy and Jacobi have successively directed 
the onward march of the mathematics. In more than fifty years of inces- 
sant thought, Euler wrote 30 separate works, and more than 700 memoirs, 
which could not altogether be contained in forty large quarto volumes. 
These writings embrace every existing branch of mathematics, and almost 
every conceivable application of them ; and there is no other mathemati- 
cian who can be placed near to him in the variety of the subjects which 
he treated. In all his writings the modern analysis is employed as the 
instrument of investigation. His " Institutiones Calculi Differentialis" was 
published in 1755, and his "Institutiones Calculi Integralis," in 3 vols., 
was published 1768-70. 

The mathematical labors of D'Alembert were chiefly directed to ques- 
tions of Astronomy and Mechanics, but in all he furnished new methods 
of analysis which contributed greatly to the improvement of the calculus. 
Among these improvements should be mentioned the method of partial 
differences, invented by him when he was studying the figures which a 
musical string assumes during its vibrations. He showed that the method 
of limits furnishes a satisfactory explanation of the theory of fluxions, 
without admitting the idea of motion, as was done by Newton. He also 
resolved the problem of the precession of the equinoxes and nutation of 
the earth's axis, and determined the quantity of each. 

The most important of the works of Lagrange is his " Mecanique Ana- 
lytique," published in 1788, of which a second and enlarged edition was 
published in 1811. This great work is entirely founded on the Calculus 
of Variations, of which he was the inventor. The whole flows from a 
single formula, and from a principle known before his time, but whose 
utility was far from being suspected. 

In 1772, Lagrange proposed to show that the theory of the development 
of functions into scries contained the true principles of the differential 
calculus, independently of the consideration of infinitely small quantities 
or limits. He called this the calculus of derived functions, and he has 
developed this view of the subject in his"Th6orie des Fonctions Analyt- 
iques," published in 1792, and at a later period in his " Lecons sur le 
Calcul des Fonctions." This system assumes for its basis the expansion 
of Taylor's theorem, which it establishes by common Algebra. 

The great work of Laplace is the ik Mecanique Celeste," being a collec- 
tion of all that had been done by himself or others concerning the theory 



THE DIFFERENTIAL AND INTEGRAL CALCULUS. 15 

of the universe. There is no branch of Physical Astronomy which is not 
materially indebted to Laplace. Superior to Euler in the power of con- 
quering analytical difficulties, he is almost his equal in the universality of 
his labors. His "Theorie des Probabilites" exhibits an analysis more re- 
fined than any hitherto used; and this work is regarded by some as 
showing even greater genius than the " Mecanique Celeste." 

The great work of Gauss is his " Theoria Motus Corporum Ccelestium," 
in which are discussed some of the most important problems in Celestial 
Mechanics, and in which the calculus is only employed as an instrument 
for the solution of problems which otherwise could not be solved. Gauss 
also gives an exposition of the method of least squares, which he invented 
prior to Legendre. 

Although Abel died at the age of 27, his brilliant discoveries had al- 
ready raised the highest expectations of the fruits of his maturer years. 
He demonstrated the impossibility of the general solution of equations of 
a higher degree than the fourth ; and his papers on Elliptic Functions 
have been honored wdth the highest praise. 

There is scarcely a branch of the mathematics, either pure or applied, 
that does not ow T e something to the labors of Cauchy ; but his fame rests 
chiefly on his residual and imaginary calculus. His published memoirs 
amount to 500, besides numerous reports on memoirs presented by others. 
These memoirs treat of the highest subjects in the mathematics — the per- 
fecting and extension of pure analysis. Among his numerous works are 
his " Course of Analysis," published in 1821 ; " Applications of the Infini- 
tesimal Calculus to Geometry," published in 1826-28 ; and his " Differen- 
tial Calculus," published in 1829. 

The chief investigations of Jacobi were connected with the theory of 
elliptic functions, and his principal work is the "Fundamenta Nova 
Theorise Functionum Ellipticarum," published in 1829 ; besides which he 
wrote many special memoirs. 

From the preceding sketch, it may be easily understood that the Dif- 
ferential and Integral Calculus, which was first exhibited by Leibnitz in a 
tract of only six pages, has in less than two centuries grown into a system 
of theorems and methods sufficient to fill a work of many quarto volumes. 
The present treatise aims only to develop the first principles of this 
science. There are numerous works in which a full exposition of the 
principles and methods of the calculus may be found, among which may 
be specially mentioned, Bertrand's "Traite" de Calcul Differentiel et de 
Calcul Integral," 2 vols.,4to. 



DIFFERENTIAL CALCULUS. 



CHAPTER I. 

DEFINITIONS. LIMITING RATIOS. DIFFERENTIATION OF ALGEBRAIC 

FUNCTIONS. 

1. Classification of quantities. In the Differential Calculus 
the quantities employed are divisible into two classes, viz., 
constants and variables. 

Constant or invariable quantities are such as retain the same 
values throughout the same discussion, and are usually denoted 
by the first letters of the alphabet, a, 6, c, etc., or by the initial 
letter of the name of the quantity. Thus in the equation to 
the circle y 2 = 2rx—x 2 , the quantity r, which represents the 
radius of the circle, undergoes no change in any one operation 
in which this equation is concerned. 

2. Variable quantities are such as may assume in the same 
discussion any value within certain limits determined by the 
nature of the problem, and are generally represented by the 
final letters of the alphabet, a?, y, z, etc. 

Thus in the equation y 2 — 2rx—x 2 , the quantities x and y ad- 
mit of different magnitudes within certain limits : x may have 
any value between and 2r ; and y may have any value be- 
tween + r and — r. 

3. Functions of one or more variables. A mathematical ex- 
pression of any form whatever involving constant and variable 
quantities is called & function of those variable quantities. 

Thus the quantities ax 2 + bx+c, V2rx—x 2 , (a + bx) n , etc., are 
all functions of x. 



18 DIFFERENTIAL CALCULUS. 

4. An explicit function is one in which the value of the 
function is directly expressed in terms of the variable and 
constants. Thus in the equation y— V%rx—x 2 , y is an explicit 
function of x. 

5. An implicit function is one in which the value of the 
function is not directly expressed in terms of the variable and 
constants. Thus in the equation x 2 — 3xy + y 2 =16, y is an im- 
plicit function of x, and x is an implicit function of y. When- 
ever the equation can be solved with respect to one or both of 
the variables involved, an implicit function may be made an 
explicit one. 

6. Notation. "When we wish to denote that y is an explicit 
function of x, without giving the precise form of the function, 
we employ the notation 

y=f¥)> y=?(x\ y=<t>( x \ etc., 

or x =f(y\ x = F(y), x = 0(y), etc. ; 

which expressions are read y is a function of x, or x is a func- 
tion of y. 

When we wish simply to denote that x and y are functions 
of each other, or that y is an implicit function of x, or x an im- 
plicit function of y, we employ the notation 

ffay) = 0, F(a,y) = 0, or 0(a?,y) = O,etc, 

which is read, function of x and y = 0. This form represents 
any equation between two variables with all the terms trans- 
posed to the first member. 

7. Variables are distinguished as independent and dependent. 
An independent variable is a quantity to which we may sup- 
pose any value to be arbitrarily assigned. A dependent varia- 
ble is a quantity whose value varies in consequence of the va- 
riation of the independent variable or variables. Thus in the 
equation to the parabola y 2 = 4:ax ) if we assign arbitrary values 






DEFINITIONS. 19 

to x, the corresponding value of y may be known from the 
equation. In this case x is made the independent variable, 
and y the dependent variable. 

This distinction is made simply for convenience, and either 
variable may be treated as the independent variable. 

8. Functions may be divided into two classes — Algebraic and 
Transcendental. An Algebraic function is one in which the 
relation between the function and its variable can be ex- 
pressed by a finite number of the ordinary operations of Alge- 
bra, viz., addition, subtraction, multiplication, division, the rais- 
ing to a known power, and the .extraction of a known root. 
Thus in the equation y=ax* + bx 2 , y is an algebraic function 
of x. 

Transcendental functions are such as can not be exhibited 
by means of algebraic expressions containing a finite number 
of terms. Infinite series and continued fractions which contain 
the variable, and can not be transformed into algebraic expres- 
sions of a finite number of terms, are transcendental functions. 
They are subdivided into exponential, logarithmic, and trigono- 
metrical. 

An exponential function is one in which the variable occurs 
as an exponent, as y—a x ^ y=z x , etc. 

A logarithmic function is one which involves the logarithm 
of a variable expression, as y— log. x, y = log. (ax+b), etc. 

A trigonometrical function is one w T hich involves trigono- 
metric operations, as ?/— sin. x, y=:t&ng.x, etc. 

9. Functions are distinguished as increasing and decreasing. 
An increasing function is one that increases when the variable 
increases. Thus in the expressions y=(a + x) 3 , y=ax 3 , etc., y 
is an increasing function of x. 

A decreasing function is one that decreases when the varia- 
ble increases. Thus in the expressions y=-, y=(a — x) 3 , etc., 
y is a decreasing function of x. 



20 DLFFEKENTIAL CALCULUS. 

A function of x may be increasing for certain values of x, 
and decreasing for other values. 

10. Limiting values of quantities.. The limit of a variable 
quantity is that value toward which it continually approaches, 
so that it can be made to differ from that value by less than 
any assignable quantity. 

For example, suppose a regular polygon to be inscribed in a 
circle. If the number of sides of the polygon be increased, 
the magnitude of each angle will increase ; and if the num- 
ber of sides be made greater than any finite number, each an- 
gle of the polygon will differ from two right angles by less 
than any assignable quantity. Two right angles is therefore 
the limit toward which each angle of the regular polygon 
continually approaches when the number of its sides is in- 
creased. 

Again, let AC be the diameter of a circle, 

P35P 

©and let any point P be taken on the circum- 
ference, from which draw PM perpendicu- 
c lar to AC. If we suppose the point P to 
move from A toward B, the line PM will 
continually increase until it becomes equal 
to BO, the radius of the circle. While the 
point P moves from B toward C, the line PM will continu- 
ally decrease until it vanishes at the point C. The radius of 
the circle is therefore the limit which the perpendicular PM 
can never exceed. 

If we convert -J- into a decimal fraction, it becomes 
0.1111, etc., 

Or To + Too" + lOoo + 10 ooo + e ^ c - 

Hence the sum of the terms of this series approaches to the 
value -g-,but can never equal it while the number of terms is 
finite. The limit of the sum of the terms of this series is 
therefore -g-. 

So also the sum of the series 

l+i + T + i + -re+ etc., 



LIMITING RATIO. 



21 



approaches nearer to 2 the greater the number of terms we 
employ ; and by taking a sufficient number of terms, the sum 
of the series may be made to differ from 2 by less than any 
quantity we may please to assign. The limit of the sum of 
the terms of this series is therefore 2. 

11. Value of a limiting ratio. When two magnitudes de- 
crease simultaneously, they may approach continually toward 
a ratio of equality or toward some other definite ratio. 

Case I. The two magnitudes may approach a ratio of equal- 
ity. Let a point P be supposed to move on 
the circumference of a circle toward a fixed 
point A. The arc AP will diminish at the 
same time with the chord AP, and by bring- 
ing the point P sufficiently near to A, we may 
obtain an arc and its chord, each of which 
shall be less than any given line, and the arc and its chord 
continually approach to a ratio of equality. The truth of this 
proposition will hereafter be proved analytically (Art. 43) ; at 
present it may be more satisfactory to the beginner to arrive 
at the same conclusion by computing the value of this ratio 
for a variety of arcs. If we begin with an arc of 60°, descend- 
ing to arcs of 30°, 10°, 5°, 1°, 1', 1", etc., and in each case com- 
pute both the length of the arc and that of its chord, and 
thence determine the ratio of the two numbers, we shall ob- 
tain the results given below : 






Chord. 


Arc. 


Ratio. 


60 degrees 


1.0000000 


1.0471976 


1.0471976 


30 " 


0.5176380 


0.5235988 


1.0115151 


10 " 


0.1743114 


0.1745329 


1.0012702 


5 " 


0.0872388 


0.0872665 


1.0003178 


1 " 


0.0174530 


0.0174533 


1.0000127 


1 minute 


0.0002909 


0.0002909 


1.0000000 



We find that the chord of one minute does not differ from the 




22 DIFFERENTIAL CALCULUS. 

corresponding arc in the first seven decimal places, and the 
chord of one second does not differ from the arc of one sec- 
ond in the first fifteen decimal places. The arc and its chord 
are therefore continually approaching to a ratio of equality. 
The same is true of an arc and its sine, or an arc and its tangent. 
The ratio to which an arc and its chord approach when the 
arc is indefinitely diminished is called their limiting ratio. 

12. Case II. Two indefinitely small magnitudes may have 
some finite ratio different from unity. Let ABO 
be a right-angled triangle, having AB equal to 
BC. Then we have AB : AC : : 1 : -/2. Let a line 
be supposed to move from BC toward A, always 
remaining perpendicular to AB, we shall have 

c the proportion AD : AE : : 1 : -\/2 ; and however 
near the line DE approaches to the point A, the ratio of AD 
to AE remains the same, even though each side of the triangle 
ADE be made less than any assigned magnitude. In like 
manner we find that the limiting ratio of a chord to half the 
corresponding arc is the ratio of 2 to 1. Two magnitudes 
therefore do not have a ratio of equality simply because each 
magnitude is indefinitely small. 

13. Caselll. The ratio of the two magnitudes may increase 
or decrease without limit. Let ABC be a tri- 
angle inscribed in a semicircle, and let AD be 
drawn perpendicular to BC. Then (Geom., 

B * c Bk. IV., Pr. 23) we have 

BC:AC::AC:DC. 
Suppose the point A approaches C, until the chord AC be- 
comes indefinitely small; DC will also become indefinitely 
small. But the above proportion still continues, and when the 
ratio of BC to AC becomes indefinitely great, the ratio of AC 
to DC will also be indefinitely great ; that is, although each of 
two quantities is indefinitely small, their ratio is indefinitely 
great. Hence we see that when two variable quantities are 




LIMITING RATIOS. 23 

supposed to diminish together indefinitely, their limiting ratio 
may have any value whatever. 

14. Increments and their limiting ratio. When, in an al- 
gebraic expression, the variable is supposed to increase, or takes 
an increment, the value of the expression undergoes a change, 
and the quantity by which its value is changed is called its 
increment. 

Thus, in the expression ax, if x receive the increment A, 
then ax will become a(x+h), and therefore the increment of 
ax is a(x + h) — ax=ah ; that is, if x be increased by A, the 
function ax will be increased by a times h. 

The increment of the independent variable is usually rep- 
resented by A, and in the following exercises, for the increment 
of x we shall write incr. x, and for the increment of y shall 
write incr. y. 

Ex. 1. If x receive the increment A, what will be the incre- 
ment of y — ax?% 

Here when x becomes x + h, the function ax 2 becomes 

a(x+h) 2 . 

Hence incr. y = a{x + A) 2 — ax 2 — 2axh + ah 2 . 

Ex. 2. What will be the limiting ratio of incr. y to incr. x in 
the last example, when A or incr. x approaches zero ? 

Dividing each side of the last equation by A, we have 

incr. y incr. y 

— 7 — or: = 2ax + ah. 

h mcr. x 

Now this equation must hold true whatever value be given to 

A. If A be taken very small, ah will also be very small ; and 

the smaller A becomes, the nearer will 2ax-\-ah approach 2ax. 

2ax is therefore the limit toward which the value of this ratio 

approaches as A diminishes. 

incr. ii 

Hence the limiting value of -: — = 2ax. 

° mcr. x 

Ex. 3. If x receive the increment A, required the increment 

of y — ax 2 -\-?>x. 

When x becomes x + A, the function y becomes 

a(x+h) 2 + 3(x+h). 



24 DIFFERENTIAL CALCULUS. 

Hence 

incr. y = a(x + h) 2 + 3(x + h) — ax 2 —3x= 2axh + ah 2 + 3h. 

Ex. 4. Required the limiting ratio of incr. y to incr. x in the 

last example. 

Dividing each side of the last equation by h, we have 

incr. y incr. y ^ 7 

\ * or -. - = 2ax+ah + 3. 

h mcr. x 

"When h approaches zero, the term ah becomes indefinitely 

small, and we have 

incr. v 

the limiting value of — =2ax+3. 

& mcr. x 

Ib.How the value of a limiting ratio is found. It will 
be seen from these examples that, in order to discover the lim- 
iting ratio of a variable to its function, we ascribe a small in- 
crement to the variable, and learn the corresponding incre- 
ment of the function. We then observe toward what limit 
the ratio of these increments approaches when the increment 
of the variable is diminished, which limit is only attained when 
the increment of the variable is supposed to be zero. 

It must not be supposed that the limiting value of a ratio is 
a mere approximation / for, while we speak of h approaching 
zero, we do this merely to aid our conception of the terms of 
the ratio ; but in finding the value of the limit of the ratio, we 
actually take h equal to zero. The limiting value of the ratio 
thus found is therefore not approximately but absolutely exact. 

Ex. 5. If x receive the increment A, required the increment 
of y = ax 3 —2bx+c. 
Inur.y = a(x-\-h) 3 — 2b(x-\-h) + c—ax 3 + 2bx—c. 

= ax 3 + 3ax 2 h + 3axh 2 + ah 3 -2bx-2bh + c—ax 3 + 2bx—c. 
= 3ax 2 h + 3axh 2 + ah 3 — 2bh. 

Ex. 6. Required the limiting ratio of the increments in the 
last example. 

Dividing eacli side of the last equation by h, we have 

incr. y incr.?/ rt ^ 7o ^ 7 

— 7-^ or : '- = 3axr + 3axh + ah 2 - 2b. 

k mcr. x 



LIMITING RATIOS. 25 

incr. ?/ 

Hence the limiting value of . — —Sax 2 — 2b. 

^ mcr. x 

Ex. 7. If the side of a square be represented by x, and it be 
increased by A, by what quantity will the surface of the square 
be increased ? Ans. 2xh + h 2 . 

Ex. 8. In Ex. 7, w T hat will be the increment given to the di- 
agonal ? Ans. h V2. 

Ex. 9. If the radius of a circle be x, and it be increased by 
h, what will be the increment of the circle ? Ans. (2xh-\rh 2 )ir. 

Ex. 10. Eequired the limiting ratio of the increments in ex- 
amples 7, 8, and 9. Ans. 2x; -y/2 ; and 2nx. 

Ex. 11. If x receive the increment A, required the incre- 
ments of the following functions : 

y — 3x 4 + h smdy=(x+l) 2 . 
A?is.l2x 3 h + 18x 2 h 2 + 12xh 3 +3h i ; and 2xh+h 2 + 2L 

Ex. 12. Eequired the limiting ratio of the increment of y to 
the increment of h in the last example. 

Ans. 12x 3 ; and2#+2. 

16. Notation adopted. It is important to have some con- 
venient notation for expressing the limiting ratio of two simul- 
taneous increments. In Ex. 4, Art. 14, it was shown that the 

incr. v 
limiting; value of . — = 2ax + 3, where y = ax 2 + 3x. Now if 

dx be put for the incr. x in the limiting value, then, in order to 
maintain a uniform notation, we must put dy for incr. y. We 
shall then have dy 

dx~ 
or, substituting the value of y, 

d(ax 2 + 3x) rt 

By -j- we therefore understand the limiting value of : — — / 
cix mcr. x 

or the limiting value of the ratio of the increment of y to the 
increment of x when the increment of x approaches zero. 
The symbol dx is called the differential of x, and dy the dif- 

B 



±=2ax + 3; 



26 DIFFERENTIAL CALCULUS. 

du 
ferential of y, and the fraction -p is called the differential co- 

dv 
efficient of y with regard to x. The process by which y- is 

obtained is called differentiation. The letter d placed before 
any algebraic expression or function indicates therefore that 
the function is to be differentiated, so that d is a symbol of an 
operation and not of a quantity. 

17. The object of the differential calcidus is to determine 

rules for finding differentials and differential coefficients, and 

to apply them to the solution of various problems both in the 

pure and mixed mathematics. 

di/ 
Not only does the symbol -j- have an important meaning, 

but a meaning may also be attached to the symbols dy and dx 
taken separately. These symbols may be regarded as repre- 
senting the simultaneous rates of increase of a function and 
of the variable upon which it depends. By the rate of in- 
crease of a variable quantity, we do not- understand its abso- 
lute increase, but rather what the increase would have been if 
it had been uniform. This distinction may be illustrated by 
the example of a falling body. Suppose a stone to fall to the 
ground from the top of a tower, and that it experiences no re- 
sistance from the air. At the commencement of its motion 
its velocity is zero, but this velocity increases uninterruptedly 
from second to second, and by the velocity at any instant we 
understand the space which would have been described in one 
second, if the velocity at that instant could be uniformly main- 
tained for one second. 

So also if y=ax 2 + 3x, and we suppose x to increase uni- 
formly, the value of y will not increase uniformly, but at each 
instant it will have a certain rate of increase depending upon 
the value of x. dy and dx are to be regarded as the simulta- 
neous rates of increase of y and x corresponding to any par- 
ticular value of x. The absolute values of dy and dx are there- 



METHOD OF DIFFERENTIATION. 27 

fore indeterminate, and they may be either finite or indefi- 
nitely small, provided their ratio is equal to the limiting value 
of the simultaneous increments of y and x. 

18. General method of differentiation. From the preced- 
ing examples and illustrations we perceive that, in order to find 
the differential coefficient of any function of a single variable, 
we give to the variable any arbitrary increment h, and find 
the corresponding value of the function ^ from which we sub- 
tract its primitive value. We then divide the remainder by 
the increment h, and obtain the limit of this ratio by making 
the increment equal to zero ; the result is the differential co- 
efficient. If the differential coefficient be multiplied by the 
differential of the variable^ it will give the differential of the 
function. 

Ex. 1. If x increase uniformly at the rate of 2 inches per 
second, at what rate does the value of the expression 2x 2 in- 
crease when x equals 6 inches ? Ans. 4S inches per second. 

Ex. 2. If x increase uniformly at the rate of 3 inches per 
second, at what rate does the value of the expression ±x 3 in- 
crease when x equals 10 inches ? Ans. 

Ex. 3. If x increase uniformly at the rate of 5 inches per 
second, at what rate does the value of the expression 2x* in- 
crease when x equals 4 inches ? Ans. 

Ex. 4. A ship sailing on a northwest course is found to be 
making north latitude at the rate of 10 miles an hour. "What 
is her rate of progress ? Ans. 14.1 miles per hour. 

19. The differential of the product of a variable quantity 
by a constant is equal to the constant multiplied by the differ- 
ential of the variable. 

Suppose we have the function 

y = ax 4: . 

Incr. y = a(x + Jif — ax 4 " — ±ax 3 h + 6ax 2 h 2 + 4:axh 3 + ah 4 ; 

incr. v 
•\ — j—^ = ±ax 3 + Qax 2 h + 4:axh 2 + ah 3 ; 



28 DIFFEKENTIAL CALCULUS. 

.\ limiting value of -. - = 4:ax 3 ; 

& mcr. x 

that is, -T-=z4;ax 3 , 

or dy—^ax 3 dx ; 

that is, the differential of ax 4 is equal to the differential of a? 4 
multiplied by a. The same process and conclusion will apply 
if y is a times any function of x. 

20. The differential of a constant term is zero ; hence a 
constant quantity connected with a function of x by the sign 
plus or minus disappears after differentiation. 

Suppose we have the function 

y=x*+b. 
Incr.y=(x + hy + b—(x* + b); 

= 4:x 3 h+6x 2 h 2 + 4,xh 3 + h*; 

/. ^l=4,x 3 + 6x' 2 h+4:xh*+ h? ; 

incr. 7j 

.-. limiting value of '— =&x 3 ; 

& mcr. x 

dy ( Q 
that is, ~ix^ x > 

where the constant term b has disappeared in the process of 
differentiation. 

21. To obtain the differential coefficient of any power of a 
variable, we must diminish the exponent of the power by unity, 
and then multiply by the primitive exponent. 

To prove this proposition let us take the function 

y=x\ 
Incr.y=(x + h) n —x n , which, by the binomial theorem, 

= nx n ~ x h + -^ — V-'A 2 + etc. ; 



incr, 



V , n(?i—l) „, 

*=n3tr x + v V ~% + etc.; 



h z 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 29 

_ . . . . - incr. y 

.•.limiting; value 01 . = nx n l : 

& mcr. x 

that is, -jr = nx n -\ 

Ex. 1. What is the differential of 5ax 3 ? Ans. 15ax 2 dx. 

Ex. 2. What is the differential of %x 2 + b ? Ans. 

Ex. 3. What is the differential of 3x 5 ? Ans. 

Ex. 4. What is the differential of 7a 2 x 6 + h 3 ? 

Ex. 5. What is the differential of ±ab 2 x 8 -c? 

Ex. 6. What is the differential of 3a 3 cx 9 -d? 

Ex. 7. If the side of an equilateral triangle increase uni- 
formly at the rate of -^ of an inch per second, at what rate is 
its perpendicular increasing when its side is equal to 8 inches ? 

Ans. —£■ inch per second. 

Ex. 8. If the side of an equilateral triangle increase uni- 
formly at the rate of \ of an inch per second, at what rate is 
the area increasing when the side becomes 8 inches ? 

Ans. 2^/3 inches per second. 

Ex. 9. If a circular plate of metal expand by heat so that 

its diameter increases uniformly at the rate of t-J-o of an inch 

per second, at what rate is its surface increasing when the 

diameter is exactly 2 inches ? tt . _ n 

Ans. 3-7T7T mch per second. 

Ex. 10. If a circular plate expand so that its area increases 
uniformly at the rate of -^ of a square inch per second, at 
what rate is its diameter increasing when the area of the cir- 
cle is exactly a square inch ? . 1 . _ _ 

Ans. gn , inch per second. 

50-V/7T r 

Ex. 11. If the diameter of a spherical soap bubble increase 

uniformly at the rate of -jV of an inch per second, at what rate 

is its capacity increasing at the moment the diameter becomes 

2 inches ? it . . 

Ans. -z inch per second. 

Ex. 12. If the capacity of a spherical soap bubble increase 



30 DIFFERENTIAL CALCULUS. 

uniformly at the rate of 2 cubic inches per second, at what 

rate is the diameter increasing at the moment it becomes 2 

inches? 1 . . 

Ans. - inch per second. 

7T x 

Ex. 13. If the diameter of a circular plate expand uniformly 

at the rate of jV of an inch per second, what is the diameter 

of the circle when its area is expanding at the rate of a square 

inch per second ? . 20 . _ 

Ans. — inches. 

7T 

Ex. 14. If the diameter of a sphere increase uniformly at the 

rate of -^ of an inch per second, what is its diameter when 

the capacity is increasing at the rate of 5 cubic inches per 

second? , 10 . . 

Ans. —r- inches. 

V7T 

Ex. 15. If the diameter of the base of a cone increase uni- 
formly at the rate of -^ inch per second, at what rate is its 
solidity increasing when the diameter of the base becomes 10 
inches, the height being constantly 1 foot ? 

Ans. 27r inches per second. 

Ex. 16. In a parabola whose equation is y 2 = 4:x, if x increase 
uniformly at the rate of 1 inch per second, at what rate does 
y increase when the abscissa equals 9 inches ? 

Ans. ^ inch per second. 

Ex. 17. In the same parabola, at what point do the abscissa 
and ordinate vary at the same rate ? Ans. When y = 2. 

Ex.18. In an ellipse whose equation is 9^ 2 +16?/ 2 =:144, at 
what point of the curve does the abscissa decrease at the same 
rate that the ordinate increases ? 

Ans. When a? =3.2, and y=1.8. 

22. The differential of the sum or difference of any number 
of functions de-pendent on the same variable is equal to the 
sum or difference of their differentials taken separately. 

Let y and z denote two functions of x, and u their sum. 
Suppose that y\ z r , u f denote the values these functions assume 
when x is changed into x + h. Then 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 31 

u = y + z, 

and u f — y t -\-z f . 

Therefore v! — it = y' ' — y + z' ' — z ; 

that is, in cr. u — incr. y + incr. z. 

_. ' . _ , incr. u incr.?/ incr. z 

Dividing by h, — jr -=- ]r+ - T -. 

This equation being true for all values of h, it will also be 
true when h approaches zero ; therefore, taking the limiting 
values by putting the ratios of the differentials for the ratios 
of the increments, we have 

du dy dz 

dx~ dx dx* 

or du — dy + dz. 

So, also, if tt — y—z^ 

we shall have da — dy—dz. 

Ex.1. What is the differential of Qx*-5x 3 — 2x ? 

Ans. (2±x 3 -l5x 2 -2)dx. 
Ex.2. What is the differential of ax} — ex? 

Arts. 
Ex. 3. What is the differential of 3ax 3 -bx^. 

Ans. 
Ex.4. What is the differential of a 6 + 3a*x 2 + 3a 2 x* + x 6 l 

Ans. 
Ex. 5. What is the differential of $x 3 -3x 2 + §x + 2 ? 

Ans. 
Ex. 6. What is the differential of 7x 5 + 6x 3 -5ax 2 + 3x-6 ? 

Ans. 
Ex. 7. In the function y=2x 3 + x 2 find the value of x when 
y is increasing 60 times as fast as x. Ans. x = 3, or — 3^-. 

23. If u represent any function of x, and we change x into 
x + h, the new value of the function to ill consist of three 
parts : 

1st. The primitive function u. 

2d. The differential coefficient of the function multiplied 
by the first power of the increment h. 



32 DIFFERENTIAL CALCULUS. 

3d. A function of x and h multiplied by the second power 
of the increment h. 

We have seen in the preceding examples that when u is a 
function of x, and we change x into x + h, the new value of 
the function consists of a series of terms which may be ar- 
ranged in the order of the ascending powers of A, and the de- 
velopment is of the following form : 

u'=u + Ah + Bh 2 + Ch 3 + etc., 
or u' = u + Ah + h 2 (B + Ch+ etc.). 

If we represent B + CA + etc., by B' where B' is a function 
both of x and h y we have 

u'=u+Ah+B'h 2 . (1) 

Transposing, and dividing by A, we find 

— j— = A+B'h, 

mcr.u 

or -. z=A+B / A; 

mcr. x ' 

.... _ . incr. u . 

.'.limiting value 01 — A* 

& mcr. x 

du . 

that is, A is the differential coefficient of the function. 

We see from equation (1) that u', the new value of the func- 
tion, consists of the primitive function u, plus the differential 
coefficient of the function multiplied by A, plus a function of 
x and h multiplied by h 2 . 

This expression for the new value of the function will be 
referred to hereafter under the form 

u f = u + Ah + Bh 2 . (2) 

It is here assumed that the function of x + h can be devel- 
oped into series in powers of h. If any function u of x + h 
shall be found that can not be developed into the form 

u' = u+Ah + Bh 2 + Ch 3 + etc., 
where the etc. includes only higher powers of h, the proposi- 
tion of this Article will not be regarded as proved for such 
function. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 33 

24. The differential of the product of two functions de- 
pendent on the same variable is equal to the sum of the prod- 
ucts obtained by multiplying each function by the differen- 
tial of the other. 

Let us designate two functions by u and v, and suppose them 

to depend on the same variable x / then, when x is increased 

so as to become x + h, the new functions, Art. 23, may be 

written u f = u + Ah + BA 2 , 

v'=v+A'h+B7i 2 . 

If we multiply together the corresponding members of these 

equations, we shall have 

u'v' = uv + Avh + Bvh 2 

+A'uh+AA f h 2 + etc., 

+ Buh 2 + etc., 

where, it will be observed, the terms omitted contain powers 

of the increment higher than A 2 . 

Transposing, and dividing by A, we have 

u'v r — uv . . , . . . 7 
t =Ay + A w+otner terms containing A. 

When we make A equal to zero, Art. 15, the terms contain- 
ing A disappear, and we have 

d(uv) . 

or, multiplying by dx, 

d{uv) = vAdx + uh!dx. 
But Adx is equal to du, 

and A!dx is equal to dv ; 

hence d(uv) = vdu + udv, (1) 

which was the proposition to be demonstrated. 

25. If we divide both members of equation (1), Art. 24, by 
uv, we shall have 

diuv) du dv 

uv ~~ U V } 

that is, the differential of the product of two functions di- 
vided by their product is equal to the sum of the quotients 

B2 



34 DIFFEKENTIAL CALCULUS. 

obtained by dividing the differential of each function by the 
function itself 

Ex. 1. "What is the differential of xy 2 % Ans. y 2 dx + 2xydy. 
Ex. 2. What is the differential of x 2 y 2 ? Ans. 
Ex. 3. What is the differential of ax 2 y 3 ? Ans. 
Ex. 4. What is the differential of ax\x 2 + 2b) ? 

Ans. 
Ex. 5. What is the differential of (x 2 + a) (2x + b) r i 

Ans. 
Ex. 6. What is the differential of (x 2 + a) (3x 2 + b) ? 

Ans. 
Ex.7. What is the differential of {a + 2x 2 ) (a-3x 3 ) + 6x 5 l 

Ans. 
Ex. 8. What is the differential of (l-^ 3 ) (l + ^ + ^ 3 ) + ^ 5 -l? 

Ans. 

26. The differential of the product of any number of func- 
tions of the same variable is equal to the sum of the products 
obtained by multiplying the differential of each function by 
the product of all the other functions. 

Let us designate three functions by u, v, and z, and suppose 
them to depend on the same variable x. Substitute y for vz, 
and we shall have uvz = uy, 

and diuvz) = d(uy). 

But, by Art. 24, d{uy) — ydu + udy ; (1) 

and since y=v& } we have, by the same Art., 

dy = zdv + vdz. 
Substituting these values of y and dy in equation (1), it be- 
comes d(uvz) = vzdu + uzdo + uvdz. (2) 

The same method is applicable to the product of four or 
more functions. 

27. If we divide both members of equation (2), Art. 26, by 
was, we shall have d{uvz) du dv dz 

uvz ~ a v z' 
which is an extension of Art 25. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 35 

Ex. 1. What is the differential of xy 2 zf 

Ans. y 2 zdx + 2xyzdy + xifdz. 
Ex. 2. What is the differential of x 2 y 2 z 3 ? 

Ans. 
Ex. 3. What is the differential of axy 3 z 3 ? 

Ans. 
Ex.4. What is the differential of x(x 2 + a)(x + 2b) < i 

Ans. 
Ex. 5. What is the differential of ax 2 (x 3 + a)(x+3b) ( l 

Aiis. 
Ex. 6. What is the differential of x\l + ax 2 )(l-ax 2 ) ? 

Ans. (2x—6a 2 x 5 )dx. 

28. To find the differential of a fraction, multiply the de- 
nominator by the differential of the numerator / subtract the 
product of the numerator by the differential of the denom- 
inator, and divide the remainder by the square of the denom- 
inator. 

u 
Let us designate the fraction by — , and suppose 

l=y; CO 

then u — vy. 

Therefore, by Art. 24, du ±= ydv + vdy ; 

whence vdy— du— ydv. (2) 

Substituting, in the second member of equation (2), the value 

of y from equation (1), we have 

, 7 udv 
vay=au— . 

Dividing by v, we obtain 

_ vdu—udv 

<ty= — -2 — ; 

. . -,/^A vdu—udv 
that is, \vj v 2 > 

which was the proposition to be demonstrated. 



36 DIFFERENTIAL CALCULUS. 

29. If the numerator of the fraction is constant, its differen- 
tial will be zero (Art. 20), and we shall have 

4-)= -4; 

\vj V 2 ' 

that is, to find the differential of a fraction whose numerator 
is constant, multiply the numerator with its sign changed by 
the differential of the denominator ', and divide by the square 
of the denominator. 

x 2 
Ex. 1. What is the differential of -^ ? 

y 

2xy 3 dx — 3x 2 y 2 dy 2xydx — 3x 2 dy 

v ~' or " ~y r ~ 

Ex. 2. What is the differential of — n ? Ans. 

x 11 

Ex. 3. What is the differential of — ~ ? Ans. 

ax 3 



x 

Ex. 4. What is the differential of z ? Ans. 

1—x 

1 + x 2 

Ex. 5. What is the differential of q ; ? Ans. 

1—x 2 

Ex. G. What is the differential of jo - 2 ? Ans. 

o — x 

Ex. 7. What is the differential of l2 i ~ ~rz~- \ % ' 

. 2xdx 
Ans. 



2' 



(* 2 -l) 

X" 

Ex. 8. What is the differential of ax 3 — -~ ? 

X 6 — x + c 

o 07 x 2 (3c—2x)dx 
Ans. 3ax 2 dx—r^ r^ • 

(x 6 —x+e) 2 

30. The differential of a function affected with any expo- 
nent whatever is the continued product of the exponent, the 
function itself \oith its exponc?it diminished by unity, and 
the differential of the function. 

This proposition is the same as Art. 21, and the demonstra- 
tion there given may be regarded as general, since the bino- 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 37 

mial theorem is true whether the exponent of the power be 
positive or negative, integral or fractional. This theorem may, 
however, be deduced directly from Art. 27. 

Let it be required to find the differential of a? w , where the 
exponent n may be either positive or negative, integral or 
fractional. 

Case I. When n is positive and a whole number. 

x n may be considered as the product of n factors each equal 
to x. Hence, by Art. 27, 

d(x n ) d(xxxx ) dx dx dx dx 

SY> n /YJ/yj/YJ^YJ /Y> ' /Y> /Y> *Y* 

t/s iL/iL/tAjtAj . • • • iAj tL> *Aj iL/ 

and since there are n equal factors in the first member of the 
equation, there will be n equal terms in the second ; hence 

d(x n ) ndx 

or d(x n ) = nx n ~ l dx. 

Case II. When n is a positive fraction. 

T 

Represent the fraction by -, and let 

S r 

u—x\ 
Raising each member of the equation to the power 5, we shall 
have u*—x\ 

and since r and s are supposed to represent entire numbers, 
we shall have, by Case I., 

su s ~ l du = rx r ~ 1 dx. 

r rs — r 

Now, since u=x% we have u s ~ l —x s . Substituting this in the 
last equation, we have 

rs — r 

sx 8 du=rx r ~ 1 dx, 
whence du — -x ° dx=-x' dx, 

T 

which is of the same form as nx n ~ l dx, substituting - for n. 

Case III. When n is negative, either integral or fractional. 
Suppose u=x~ n , 



38 DIFFEKENTTAL CALCULUS. 

which may be written u=— . 

Differentiating by Art. 29, we have 

7 — d(x n ) 

and differentiating the numerator by Case I., or by Case II. if 
n represent a fraction, we have 

nx n ~ l dx 
x 2n » 
or, reducing, we have 

du——nx~ n ~ x dx^ 
which is of the same form as nx n ~ l dx, by substituting —n for 
+ n. 

The proposition above enunciated may therefore be consid- 
ered general, whatever be the exponent of x. 
Ex. 1. What is the differential of ax^ 1 ? 

Ans. a(n + l)x n dx. 
ci — 
Ex. 2. What is the differential of T x n + c? Ans. 

o 

2 

Ex. 3. What is the differential of ab 2 x* % Ans. 

Ex. 4. What is the differential of bx 5 ? Ans. 

Ex. 5. What is the differential of cx~ 3 ? Ans. 

i 
Ex. 6. What is the differential of a?y 2 z 3 ? Ans. 

Ex. 7. If the area of a square increase uniformly at the rate 
of tV of a square inch per second, at what rate is the side in- 
creasing when the area is 100 square inches ? Ans. 

Ex. 8. If the solidity of a cube increase uniformly at the 
rate of a cubic inch per second, at what rate is the edge in- 
creasing when the solid becomes a cubic foot ? Ans. 

31. The differential of the square root of a variable quan- 
tity is equal to the differential of that quantity divided by 
twice the radical. 

Let it be required to find the differential of 



Vx or 



x 



,7 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 39 

According to the preceding article, 

1 1 *-] 
d(x*) = nX* dx, 

dx 

The preceding proposition is only a particular case of Art. 30 ; 
but we have so frequent occasion to differentiate radicals of the 
second degree, that a special rule for this case is desirable. 

Ex. 1. What is the differential of Vax? ? 

3ax 2 dx ii 

Ans. , or %a 2 x 2 ax. 

^Vax 3 

Ex. 2. What is the differential of Vabx 1 ? Ans. 

Ex. 3. What is the differential of Vax 5 ? Ans. 

x 
Ex. 4. What is the differential of a-y/x — t> ? 

Ans. 
Ex. 5. What is the differential of Vax + Vc 2 x 3 ? 

Ans. 



oy 



,2 



Ex. 6. What is the differential of , ? 



(l+O^ 



32. Polynomial raised to any power. Art. 30 extends to 
any function of x affected with any exponent. Let us sup- 
pose this function to be a polynomial, and let it be required to 
differentiate the function 

u—{ax+x 2 ) n . 
Substitute y for ax+x 2 , and we have 

u = y n . 
Whence, by Art. 30, du — ny n ~ l dy. 
Restoring the value of y, we have 

du — n(ax + x 2 ) H ~ l d(ax + x 2 ). 



40 DIFFERENTIAL CALCULUS. 

From which we see that to obtain the differential of a polyno- 
mial raised to any power, we must diminish the exponent of 
the power by unity, and then multiply by the primitive expo- 
nent, and by the differential of the polynomial. 

In the preceding example the differentiation of ax + x 2 is 
only indicated. If we actually perform the differentiation, we 
shall have 

du = n(ax + # 2 ) n_1 (a + 2x)dx. 

Ex. 1. What is the differential of Va+bx 2 % 

bxdx 
Ans. . 

va + bx 2 

Ex. 2. What is the differential of (ax 2 + x 3 ) 3 ? 

Ans. 



Ex. 3. What is the differential of Vax+bx 2 +cx 3 ? 

Ans. 
Ex. 4. What is the differential of (ax-x 2 ) 5 ? 

Ans. 

Ex. 5. What is the differential of {a+bx 2 f ? 

Ans. 

\_ 

Ex. 6. What is the differential of (a+x 2 ) n ? 

Ans. 

Ex. 7. If x increase uniformly at the rate of -j-J-^ of an inch 
per second, at what rate is the expression (1 + x) 3 increasing 
when x equals 9 inches ? Ans. 

Ex. 8. A boy standing on the top of a tower whose height 
is 60 feet, observed another boy running toward the foot of the 
tower at the rate of five miles an hour on the horizontal plane ; 
at what rate is he approaching the first when he is 80 feet from 
the foot of the tower % Ans. 4 miles an hour. 

Ex. 9. In the last example how far was the boy from the 
foot of the tower when he was approaching the foot twice as 
fast as he approached the top ? Ans. 20 -\/3. 

33. To differentiate a function of a function. Let ^bea 
function of y, and y a different function of x ; that is, 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 41 

u=f(y) (i) 

and y=0O»); (2) 

it is required to find the differential coefficient of u regarded 

as a function of x. 

In the equation y = <j>(%) let x be increased by h ; then y 
will receive an increment y'—y, which we will denote by Jc. 
When we substitute x+k for x in equation (2), we have (Art. 
23) y f = y+Ah+Bh 2 + etc. ; 

or V -^- = A + BA + etc., (3) 

where A is equal to -r-. 

Also, when we substitute y+Jc for y in equation (1), we have 
u'=u+Wc+NJc 2 + etc.; 

or , =M+N&+ etc., (4) 

where M is equal to -r~. 

Multiplying together equations (3) and (4), member by mem- 
ber, we have 

^^ x lj^ = AM+A'NJc+BWi + etc. 

But since, by hypothesis, y f — y — Jc^ we have 

xi f —u incr. u 4 __ 4 ___ _„- 

— t— , or = — — =AM+AN£+BMA+ etc. 

ft 11101 . X 

Now when h is made^O, Jc will also = 0, since & denotes the 
increment of y caused by ascribing the increment h to x. 
Hence for the limiting ratio we shall have 

ax 

But A = ^ and M=-t-; hence 

cfo^ du dy 

dx dy dx y 

that is, the differential coefficient qfu, regarded as a function 



42 DIFFERENTIAL CALCULUS. 

ofx, is equal to the differential coefficient ofu regarded as a 
function of y, multiplied by the differential coefficient of y 
regarded as a function ofx. 

du 
Ex. 1. Given u—3y 2 and y=x 3 +ax 2 to find y. 

du _ _ dy n _ rt 
-y- = 6y and ~^-=3x 2 + 2ax; 

du 
hence ^ = 6y(3^ 2 + 2aaj) = G(x 3 + ax 2 )(3x 2 + 2ax). 

Ex. 2. Differentiate the function ^= Va 2 — x 2 . 

Assume y=a 2 —x 2 \ we then have the two functions 

i 
u=y 2 and y=a 2 —x 2 ; 

. du —x 

hence 



flfe -vV— x 2 
Ex. 3. Differentiate the function ii~(a+bx m ) n . 

Ans. -j- — hnnx m - l (a+bx m ) n ~\ 

Ex. 4. Differentiate the function w=(a+ -/a?) 3 . 

du 3(a+Vx) 2 
dx~~ 2*\/x 

34. The preceding principles will enable us to differentiate 
all algebraic functions of one independent variable, whether 
explicit or implicit, however complicated they may be, because 
all the combinations of quantities that can be formed by the 
common operations of Algebra have been considered in the 
demonstrations. 

(1 -4- T* 

Ex. 1. What is the differential of the function u — — ; — = ? 

a+x 2 

According to Art. 28, 

, _(a + x 2 )dx— 2x(a+ x)dx 

du=z (a+x 2 ) 2 

which may be reduced to 

T (a— 2ax— x 2 )dx 

du= - — ■ — ^ — . 

(a+x 2 ) 2 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 43 



Ex. 2. Differentiate the function u=Vx 2 + a-\/x. 

Ans. 
(b+x) 2 



Ex. 3. Differentiate the fanction u= 



Ex. 4. Differentiate the function u- 



x 

Ans. 

x 2 



Ex. 5. Differentiate the function u — 



{a + x 3 ) 2 ' 
Ans. 
1 



~(a + x 2 ) n ' 

Ans. 
Ex. 6. What is the differential of the function 
u — x{a + x)(a 2 + x 2 ) ? 

Ans. du = (a 3 + 2a 2 x + 3ax 2 + 4:X 3 )dx. 

Ex. 7. What is the differential of the function 

u = (a + bx) 2 (?n + nx) 3 ? 
Ans. du = 3?i(a + bx) 2 (in + ?ix) 2 dx + 2 b(a + bx)(m + ?ix) 3 dx. 

Ex. 8. What is the differential of the function 

u = (a+bx 2 ) 3 (c+ex*) 5 l 
Ans. du = 20(a + bx 2 ) 3 (c + ex^fex 3 dx + Gb(c + ex*)\a + bx 2 ) 2 xdx. 

Ex. 9. What is the differential of the function 

u=\/4:x 3 -51 

. _ 6x 2 dx 
Ans. du- 



V±c 3 -5 
Ex. 10. What is the differential of the function 

u = xia 2 + x 2 ) Va 2 — x 2 ? 

(a*+a 2 x 2 — 4:x*)dx 

Ans. du — , — . 

Ya 2 —x 2 

Ex. 11. What is the differential of the function 

a+x 



U- 



? 



Va— x 

3a— x T 
Ans. du— zdx. 

2{a-xf 



44 DIFFERENTIAL CALCULUS. 

Ex. 12. What is the differential of the function 

- g » 

, dx 

Arts. au=- 



U- 



Vi-^+M 1 -^)' 

Ex. 13. What is the differential of the function 

4-y/a? _ 

Ans. du -(z +x y^ x - 

Ex. 14. What is the differential of the function 

a 2 —x 2 

u=z a* + a 2 x 2 +x*' 

-<2 x(2a*+2a 2 x 2 -x*)dx 
(a* + a 2 x 2 +x*) 
Ex. 15. What is the differential of the function 

x n 



Ans. du~- 



U ~{l + xf 



. 7 . TtX C&X 

Ans. au=- 



"(l+a?)-+ 1- 

Ex. 16. What is the differential of the function 

u—(a— x)Va 2 + x 2 ? 

. _ (a 3 — ax+2x 2 )dx 

Ans. du= — , . 

Va 2 +x 2 

Ex. 17. What is the differential of the function 

u=(a 2 —x 2 )Va-+x'} 

1 i 

Ans. du=-^(a — 5x)(a+x) 2 dx. 

Ex. 18. What is the differential of the function 



u = 



ay . 



vx +y axydx—ax 2 dy 
Ans. du—— g- 2 -. 

(x 2 + y y 

Ex. 19. What is the differential of the function 

u = {2a 2 + 3x 2 )(a 2 -xyi 

i 
Ans. dx = — 1 5x 3 (a 2 — x 2 )^dx. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 45 

Ex. 20. What is the differential of the function 



Va + x+Va — x n 

<*= j 7= ? 

va+x— V a— x 



a 2 + aVa 2 —x 2 _ 

Ans. au=— . ax. 

x 2 Ya 2 —x 2 

Ex. 21. What is the differential of the function 

a+%bx . 

( a + bx ) 2 A J -% 2 xdx 



46 DIFFERENTIAL CALCULUS. 



CHAPTEK II. 

DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 

35. A transcendental function (Art. 8) is one in which the 
relation between the function and its variable can not be ex- 
hibited by means of algebraic expressions containing a finite 
number of terms. 

Thus a x y a log. x, sin. x, tang, x, etc., are transcendental func- 
tions ; the first is called an exponential function, the second a 
logarithmic function, while the third and fourth are trigono- 
metrical functions. 

36. To find the differential of a logarithm. Let it be re- 
quired to differentiate y— log. x. 

It is shown in Algebra, Art. 425, that 

log. (l + m)=M(m— g-f-g— etc.), 

where M represents the modulus of the system of logarithms, 
and m represents any quantity whatever. Substituting for m 

. h 

the quantity -, we obtain 

x 

log-(— j=M(--^- 2 + ^3-etc.j. 

But since the logarithm of a quotient is equal to the logarithm 
of the dividend diminished by the logarithm of the divisor, we 

have , , n i ,«/* h2 *' \ 

log. (x+h)-hg. *= M (^- 2^2+3^3- etc -J; 

and dividing by h, we have 

log, (x + //■) -log, x _ M /l _^_ 
h 



j __/l h h? \ 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 47 



which is a general expression for the ratio of the increment of 
the function to that of the variable ; that is 

incr. 

incr. 
Hence the limiting value of 



ir.y /I h_ h?__ \ 
iv.x~ \x~~2x 2 3x 3 ~~ j 

of 

incr. y M 



incr. x~~ x ' 

. dy d log. x M 

' dx~ dx ~~ x J 

Mdx 
or dlog.x=—^-; 

that is, the differenticd of the logarithm of a quantity is equal 
to the modidus of the system into the differential of the quan- 
tity divided by the quantity. 

37. Throughout this treatise, Naperian logarithms will always 

be used, unless the contrary is stated. Denoting the Naperian 

logarithm by log., and remembering that the modulus of this 

system is unity (Algebra, Art. 426), we have 

_. dx 

dlog.%=—; 

that is, the differential of the JVaperian logarithm of any quan- 
tity is equal to the differential of the quantity divided by the 
quantity. 

Ex. 1. Differentiate the function 2/= log. (ax 3 ). 

3dx 

Ans. dy— . 

* x 

Ex. 2. Differentiate the function y=log. (a+x). 

a 7 dx 
Ans. dy—- 



a + x 
Ex. 3. Differentiate the function y— log. (a + x 2 ). 

. .. %xdx 

Ans. dy— 5. 

* a + x 2 

Ex. 4. Differentiate the function y — log.(ax n ). 

_ ndx 

Ans. dy= . 

x 



48 DIFFEKENTIAL CALCULUS. 

X 

Ex. 5, Differentiate the function y— log. . 

Ya 2 +x 2 

4 -w & CLX 

Arts, ay— , 2 , — =r. 
* a?(# z + #) 

ci -\- 1* 
Ex. 6. Differentiate the function y=\og. . 

_ 2adx 

Arts, dy—-^ 5. 

J a 2 —x 2 

Ex. 7. Differentiate the function y— log. \x+(x 2 — l)" 2 ). 

. y CVX 

Am. ay=—= . 



y '# 2 +l_l 
Ex. 8. Differentiate the function v^log. . — . 

y B vV + 1 + 1 



Vx 2 + 1 



Ex. 9. Differentiate the function y=x 3 — 3 log. (1+cc 3 )" 3 ". 

. _ 3^ 5 &? 
* l + # 3 

o<» loo* HP 

Ex. 10. Differentiate the function y——\ — ^— + log.(l— x). 

. 7 log. a? . rfte 
y (1— a?) 2 
Ex. 11. Differentiate the function y=#(log. #) w . 

Ex. 12. Differentiate the function 7= log. \ , \ . 

* 5 U+Va 2 + x 2 ) 

7 a&? 



^Va 2 +^ 2 

38. When a function consists of products and quotients of 
roots and powers, it is frequently advantageous to take the 
logarithm of each member of the equation before differenti- 
ating. 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 4:9 

Ex. 1. Differentiate the function y=(a+x) m (b + x) n . 

Taking the logarithm of each member, we have 

log. y = m log. (a + x) + n log. (b + x). 

XT , , _^ dy mdx ndx 

Hence (Art. 37) — =— — + ft ~ • 

. y y a+# b + x' 

therefore dy = (a + x)'\b + ®)"h^ + fc^) dx - 

(x + 4) 2 

Ex. 2. Differentiate the function y = ■£-. 

y x + 2 

x n 

Ex. 3. Differentiate the function y—-r\ w. 

* (l+x) n 

. 7 nx n ~ l dx 
Ans. dy—-. 



i 

\"2" 



"(l+#)" +1 

(#— 1Y^ 
— -T I . 

-4.WS. <%=: 1 5". 

(^-1)^ + 1)^ 

Ex. 5. Differentiate the function y=x(a 2 + x 2 )Va 2 — x 2 . 

a* + a 2 x 2 —4:x* 7 
Ans. dy— \ — dx. 

(a 2 -xy 

3 /l+X 

Ex. 6. Differentiate the function y — x^\/ = . 

j. — x 

% + ^x-Zx 2 I x _ 
* 2(1 — a?) v 1— # 2 

39. Illustration from logarithmic tables.. The differential 
coefficient of a function denotes the limiting ratio of the incre- 
ment of the function to that of the variable. It therefore de- 
notes very nearly the actual ratio of the two increments when 
the increment of x is taken very small. This principle is illus- 
trated by a table of common logarithms. What is called the 

C 



50 DIFFERENTIAL CALCULUS. 

tabular difference is the increment of the logarithm corre- 
sponding to an increment of unity in the natural number. It 
represents, therefore, very nearly the differential of the loga- 
rithm. This will be understood from a few examples. 

Ex. 1. The common logarithm of 4825 is 3.683497 ; what is 
the logarithm of 4826, assuming that the tabular difference is 
the differential coefficient of the logarithm ? 

Let x represent any number, and y its logarithm ; then 

y=log.x. 

Hence (Art. 35) dy— . 

x 

But M, the modulus of the common system, is 0.434294; 

0.434294 
hence dy = \ " — dx = 0.0000906?x. 

The increment of the logarithm is therefore 0.000090 for an 
increment of unity in the corresponding natural number; 
hence the logarithm of 4826 is 3.683587, and this is the loga- 
rithm found in the table. 

Ex. 2. Required the tabular difference for numbers between 
9651 and 9652. 

Ex. 3. Required the tabular difference for numbers between 
5791 and 5792. 

Ex. 4. Required the tabular difference for numbers between 
3810 and 3811. 

40. To differentiate an exponential function. Let a repre- 
sent a constant quantity, and let 

y=a*. 
Taking the logarithm of both members, we have 

log. y~x log. a. 
Differentiating both members, we have (Art. 37) 

— =zdx log;, a. 
y & 

Therefore dy = y log. adx = a* log. adx ; 

that is, the differential of a constant quantity raised to a 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 51 

power denoted by a variable exponent^ is equal to the power 

multiplied by the Naperian logarithm of the constant into 

the differential of the exponent. 

Ex. 1. Differentiate the function y — a^. 

Ans. dy=na nx log. a.dx. 

Ex. 2. Differentiate the function y — x n a x . 

Ans. dy = x n ~ l a x (n + x log. a)dx. 

Ex. 3. Differentiate the function y — a x {x— 1). 

Ans. dy — a x (x log. a— log. a + l)dx. 

a x 

Ex. 4. Differentiate the function y—- . 

9 1 + x 

A?7 „ 7 ar(x log. a +log .a-l)dx 
Ans. dy- (1 + a?)2 

Ex. 5. Differentiate the function y=(a x + x) 2 . 

Ans. dy=2(a x + x)(a x log. a + l)dx. 

a x — \ 
Ex. 6. Differentiate the function y —-—-—, 

* a x + l 

7 2a x log. a.dx 

Ans - d «= {a'+iy ■ 

41. To differentiate a variable quantity with a variable ex- 
ponent. 

Let u — y x . 

Taking the logarithm of each member, we have 

log. ii — x log. y. 

Differentiating each member, we have (Arts. 24 and 37) 

du . , xdy 
— = lo^. ydx + , 

u & y y 

• _ _ uxdy 
or du = u log. ydx + -. 

Substituting the value of u, we have 

d(y x )=y x log.ydx+y*x^, 

= y* log. ydx+xy x ~ l dy; 
that is, to differentiate a quantity formed by raising a variable 
quantity to a power whose exponent is variable, differentiate 



52 DIFFERENTIAL CALCULUS. 

first as though the root alone were variable, then as though 
the exponent alone were variable, and take the sum of the re- 
sults. 

fl x / f/\ x 
Ex. 1. Differentiate the function u=— x or I - J . 

Ans. du=l-\ (log. - — l)dx. 

Ex. 2. Differentiate the function u=x*. 

. _ x x (l — log. a?) T 
Ans. au= r- 2 — -ax. 

x 2 

(x\ nx 
nj 

Ans. du=nl-j <l+log. — >dx. 

Ex. 4. Differentiate the function u — {-\/x) x . 

X 

Ans. du—-^{l+\og.x)dx. 

Ex. 5. Differentiate the function u~a x , in which the varia- 
ble exponent is x x . 

Ans. a x x x (l +log. x) log. #&?. 

DIFFERENTIALS OF TRIGONOMETRICAL FUNCTIONS. 

42. Arc and sine compared. Let AB be an arc of a circle 
^ whose radius is AC. Take the arc AB 7 
equal to AB. Join BB 7 , and draw DD 7 
touching the circle at A. Then BE is the 
L sine and DA the tangent of the arc AB. 

The chord BB 7 , being a straight line, is 
D ' shorter than the arc BAB 7 ; therefore the 
sine BE, which is half of BB 7 , is less than the arc BA, which 
is half of BAB 7 . Therefore the sine is less than the arc. 
Again, the area of the sector ABC is measured by 
-^ACxarc AB ; 




DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 53 

also the area of the triangle ADC is measured by 

^ACxAD. 
But the sector ABC is less than the triangle ADC, being con- 
tained within it ; hence 

| AC x arc AB <1AC x AD. 
Consequently, arc AB<AD; 

that is, the arc is less than its tangent; or an arc of a circle 
(not exceeding a quadrant) is greater than its sine and less 
than its tangent. 

If we put h for the arc AB, we shall have 
sin. A<A<tang. h. 

43. Limiting ratio of an arc to its sine. By Trigonometry 

(Art. 28), 

sin. h cos. h 

tang. h~~ R 

But the cosine of zero is equal to radius, hence the limit of the 

value of 

sin. h . 

7 is unity ; 

tang, a J ' 

and since the arc is comprised between the sine and tangent, 

the limit of the value of 

sin. h . 

—j^- is unity; 

that is, the limit of the ratio of an arc to its sine is unity. 
So also the limit of the ratio of an arc to its tangent or its 
chord is unity. 

44. To find the differential of sine x. Let ^=sin. x. 
If we increase x by A, then 

u' = sm. (x+h), 

and u'—u= sin. (x + h) — sin. x. (1) 

By Trigonometry, Art. 75, 

2 
sin. a— sin. b — ~^ sin. £(a— b) cos. i(a+b). (2) 



54 DIFFERENTIAL CALCULUS. 

Making a—x+h and £=#, equation (2) becomes 

2 
sin. (x + h) — sin. x = — sin. \h cos. (x + ^h). 

Hence equation (1) becomes 

2 
v! — u=— sin. ^h cos. (#+-|/A. 

Dividing both members by A, and both terms of the fraction 

in the second member by 2, we have 

u f — u sin . \h cos. (x + ^h) 

h = \h X R ; 

incr. w sin. -|A cos. fe+4A) 

that is, i = 17 2 X V a ^ 

incr. a? *£ R 

Now at the limit (Art. 43), E£lE_ = l. 

2™ 

_ $w cos. a? 

Hence, -r— = _ , 

5 da? R ' 

_ 7 . cos. #d# 

or du — d&m.x= — — . 

K 

Radius is usually taken as unity, in which case we have 

d sin. a?:=cos. xdx. 

45. To find the differential of cosine x. Let ^ = cos. x. 

Then du = d cos. x~d sin. (90°— x). (1) 

But by Art. 44, 

, . ,™o x cos.(90°-x)d(90°-x) 
d sin. (90° -a?) = ^— ^ * 

But cos. (90° — x) = sin. a?, 

and d(90°—x)=—dx. 

Hence by substitution equation (1) becomes 

7 sin. xdx 
d cos. x— — — — . 

_Lv 

Also, since the versed sine of an arc, less than ninety de- 
grees, is equal to radius minus the cosine, we have 

_ , , 7 / - r ^ . sin. xctx 

d versed sin. x — d (R— cos. x) — — — — . 

R 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 55 

46. To find the differential of tangent x. Let ^=tang. x. 
bmce tang. a? = ( Ing., Art. 28), we have 

COS. X 

, Ecos. xd sin. cc — E sin.xd cos. x 
a tang;. a?= « , 

° COS. a? ' 

(cos. 2 a?+sin. 2 a?)<fe 

cos. 2 a? 

But cos. 2 a? + sin. 2 a? = E 2 . 

, , W'dx sec. 2 xdx 

Hence d tans:, x = 5— = — ™ — . 

cos/a? Iv 3 

47. To find the differential of cotangent x. Let ^=cot. & 
Then du=d cot. x=d tang. (90°-- x). (1) 
But by Art. 46, 

tf tang. (90°— a?) = —7^ {. 

b v y cos. 2 (90°— x) 

Also, t?(90°— a?)= — dx, 

and cos. 2 (90° — a?) = sin. 2 a?. 

Hence by substitution equation (1) becomes 

_ H 2 dx cosec. 2 xdx 

d cot. x—- 



sin/aj 



E 2 



48. To find the differential of secant x. Let ^=sec. a?. 

E 2 

Since sec. a?= , we have 

cos. ar 

T E 2 ^ cos. a? E sin. xdx tans:, a? sec. xdx 

a sec. a?= — « = 9 = — — ™ • 

cos.a? cos. 2 x E 2 

49. Tofi?id the differential of cosecant x. Let ^= cosec. x. 

Then du—d cosec. x — d sec. (90°— a?). 

-r, t , ,.- % cot. a? cosec. a%? (90°— a?) 
But # sec. (90 — x) = ^2 — • 

TX _ cot. x cosec. xdx 
Hence d cosec. a?= — ™ • 



56 DIFFERENTIAL CALCULUS. 

Ex. 1. In what part of the quadrant does the arc increase 
twice as fast as its sine ? 

Let y~ sin. x y then dy= cos. xdx; that is, the rate of increase 
of the sine is equal to the rate of increase of the arc multi- 
plied by cos. x. Now cos. 60°=^-. 

Hence at 60° the arc is increasing twice as fast as its 
sine. 

Ex. 2. In what part of the quadrant does the arc increase 
three times as fast as its sine ? Ans. At 70° 32'. 

Ex. 3. In what part of the quadrant does the arc increase 
half as fast as its tangent ? Ans. At 45°. 

Ex. 4. The natural sine of 30° is 0.500000; what is the sine 

of 30° v% 

The length of an arc of V is found by dividing 3.14159 by 
the number of minutes in 180°, and we obtain 0.0002909, 
which represents the increment of the arc. The increment of 
its sine will be found by multiplying this number by cos. 30°, 
or 0.866, and we obtain 0.000252, which is the tabular differ- 
ence corresponding to the natural sine of 30°. Hence the sine 
of 30° V is 0.500252. 

Ex. 5. The natural sine of 10° 30' is 0.182236 ; what is the 
sine of 10° 31' ? Ans. 0.182522. 

Ex. 6. The natural cosine of 65° 10' is 0.419980 ; what is 
the cosine of 65° 11' ? Ans. 0.419716. 

Ex. 7. The natural tangent of 45° is 1.000000 ; what is the 
tangent of 45° 1' I Ans. 1.000582. 

Ex. 8. The natural cotangent of 60° is 0.577350 ; what is 
the cotangent of 60° V ? Ans. 0.576962. 

Ex. 9. Differentiate the function y=$m. mx. 

Ans. dy — cos. mxd{rax) — ^n cos. mxdx. 

Ex. 10. Differentiate the function ^=sin. 3 #. 

Ans. dy=3 sin. 2 xd sin. x=3 sin. 2 x cos. xdx. 

Ex. 11. Differentiate the function y — sin. n x. 

Ans. n sin. n-1 # cos. xdx. 

Ex. 12. Differentiate the function y = a*' m ' x . 

Ans. dy=a* inx log. a cos. xdx. 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 



57 



Ex. 13, 
Ex. 14 

Ex. 15. 
Ex. 16. 
Ex. 17. 



Differentiate the function y=(cos. x) n . 

Ans. dy~ — n (cos. x) n ~ l sin. xdx. 
Differentiate the function y— sin. 2x cos. x. 
dy 



Ans. 



dx~ 



:2 cos. 2x cos. x— sin. x sin. 2x. 



=:cos. 2# cos. # + cos. 3#. 
Differentiate the function y=x— sin. a? cos. a?. 

^4^5. dy=2 &m. 2 xdx. 
Differentiate the function y— tang, a?— x. 

Ans. dy=:tmg. 2 xdx. 
Differentiate the function y—t2iT\g. n x. 



Ans. dy- 



n tang." 



tL(Jb<As 



cos. 2 # 



Ex. 18. Differentiate the function y= Vtang. 2x. 

. _ sec 2 2xdx 
Ans. ay =—7= 

V tang. 2x 

Ex. 19. Differentiate the function y=cot. 2 x. 

. _ 2 cot. a%?# 
Ans. dy= — — ; — « . 

Ex. 20. Differentiate the function y = 3 tang. # + (tang. a?) 3 . 

Ans. dy=3(sec. xfdx. 

50. Geometrical illustration. The values found in Arts. 
44-49 for the differentials of sin. x, cos. x, 
etc., may be deduced from the principles 
of elementary geometry. 

Let AB be an arc of a circle whose cen- 
tre is O ; let BM be its sine, OM its co- 
sine, AE its tangent, and OE its secant, 
and let the radius OA be taken as unity. 
Let the arc AB be represented by x, and 
suppose x to receive a small increment 
BC. Draw BD perpendicular to CN, and 
with radius OE draw the circular arc EG ; then CD will rep- 
resent the increment of sin. x, BD taken negatively will repre- 
sent the increment of cos. x y EF will represent the increment 

C2 




58 DIFFEEENTIAL CALCULUS. 

of tang, x, and GF the increment of sec. x. Suppose the 
chords BC and EG- to be drawn, forming the rectilinear tri- 
angles BCD and EFG, then, when the arc BC is indefinitely 
diminished, since the limiting value of the angle OBC is 90°, 
the limit of the ratio of OBC to MBD is unity ; that is, the 
limit of the ratio of DBC to MBO is unity. So also the limit 
of the ratio of BCD to BOM is unity; that is, the triangle 
BCD in its limiting form is similar to the triangle BOM or 
EOA. For a like reason, the triangle EGF in its limiting 
form is similar to the triangle OMB or OAE. Hence we have 
the following equations : 

Ihe limit oi t^t or — - — is -r— or cos. x. 
CJ3 ax OB 

m, i. .. o BD d cos. a? . BM 

The limit of -^rzz or — is -przz or sin. x. 

CB ax OB 

Also, by similar figures, 

EG EO EO_ 
BC"~BO~-AO" SeC ^ ; 

,,,..,, EF . EO 

and the limit of ^77 is -— or sec. x. 
EG AO 

Hence the limit of 

EF , . P EF EG dtsmg.x 

— , that is, of — x — or - -^— =«e.* 

FG AE 

So also the limit of ^77 is —7- or tang. x. 
EG AO ° 

Hence the limit of 

FG , . JG EG dsec.x 

_, that ib, of — x w or -^-=tang. x see. x. 

51. Differentials of logarithmic sine, cosine, etc. By com- 
bining the results of Arts. 44-4:9 with the rule in Art. 36, we 
may differentiate the logarithm of any trigonometrical func- 
tion. 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 59 

By Art. 36, calling radius unity, we have 

. 71 . Md sin. x M cos. xdx ,, 

1. a loo;, sm. x= — = = : = M cot. xdx. 

° sm. x sm. x 

_ .. 1 Mrf cos. as M sin. xdx _ _ 

2. <z log. cos. a?= = — = — M tans:, xdx. 

& cos. x cos. a? to 

_ , . Md tans;, a? Mdx Mdx 

3. a loo;, tans;, x— — — = « — : —~ 

& tang, x cos/a? tang, a? sm. a? cos. a? 

2Mdx 

~~sin. 2a?' 
M# cot. x Mdx Mdx 



4. 6? log. cot. x— 



cot. a? sm. 2 a? cot. x sin. a? cos. x 

%Mdx 



sin. 2af 

Hence the differentials of the logarithmic tangent and co- 
tangent of any arc differ only in sign. 

It may also be easily proved that the differential of the log- 
arithmic tangent is equal to the arithmetical sum of the dif- 
ferentials of the sine and cosine of the same arc. 

Ex. 1. In what part of the quadrant does the arc increase at 
the same rate as its logarithmic sine ? 

We have found d log. sin. x—M cot. xdx; 
that is, the rate of increase of the logarithmic sine is equal to 
the rate of increase of the arc multiplied by M cot. x, or 

J— Now M = 0.434294, and tang. 23° 28' = 0.434294. 
tang, x ' fe 

Hence at 23° 28' the arc and its logarithmic sine are increas- 
ing at the same rate. 

Ex. 2. In what part of the quadrant does the arc increase at 
the same rate as its logarithmic tangent ? Ans. 30° 9'. 

Ex. 3. The logarithmic sine of 30° 5' is 9.700062; what is 
the logarithmic sine of 30° 6' ? 

The increment of x found in Ex. 4, Art. 49 is 0.0002909, 
which, multiplied by M, gives Mdx= 0.0001263. 

Multiply this result by cot. 30° 5', and we have d log. sin. x 
= 0.000218. 



60 DIFFERENTIAL CALCULUS. 

Hence the logarithmic sine of 30° 6' is 9.700280. 

Ex. 4. The logarithmic cosine of 45° 35' is 9.845018 ; what 
is the logarithmic cosine of 45° 36' ? Arts. 9.844889. 

Ex. 5. The logarithmic tangent of 46° 30' is 10.022750 ; 
what is the logarithmic tangent of 46° 31 ' % 

Ans. 10.023003. 

Ex. 6. The logarithmic cotangent of 50° 15' is 9.919962 ; 
what is the logarithmic cotangent of 50° 16' ? Am. 9.919705. 

In the following examples Naperian logarithms are sup- 
posed to be used : 
,Ex. 7. Differentiate the function y=\og. sin. 2x. 

d sin. 2x 2 cos. 2xdx _ _ _ 

Ans. ay——. — ~ — = — ; — ^ =2 cot. 2xdx. 

a sm. 2x sin. 2a? 

Ex. 8. Differentiate the function y = log. sin. 2 #. 

Ans. dy—2 cot. xdx. 
Ex. 9. Differentiate the function 



y=log.v / + ^' ^ 41og.(l + sin. a?)— Jlog.(l-sin. x). 

J- — Sill. X 

d(l-\-$m.x) d(l — sm.x) dx 

Ans. dy—\—r—-. -— \-\ = -= . 

* J 1 + sm. x z 1 — sm. x cos. x 

Ex. 10. Differentiate the function ?/ = log.V / sin. x. 

Ans. dy=^ cot. xdx. 



Ex. 11. Differentiate the function y=log.v/y 



— cos. x 



+ cos. x 

. 7 dx 
Ans. dy— — 



sm. x 



Ex. 12. Differentiate the f unction y — log A/sin. x+ log. Vcos. x. 

Ans. dy—- 



"tang. 2x 



INVERSE TRIGONOMETRICAL FUNCTIONS. 



52, Notation adopted. In the preceding articles we have 
found the differentials of the sine, cosine, etc., in terms of the 
arc as an independent variable. It is sometimes more con- 
venient to regard the arc as the function, and the sine, cosine, 



INVERSE TRIGONOMETRICAL FUNCTIONS. 61 

etc., as the independent variable. Such functions are called 
inverse trigonometrical functions, and they are commonly 
designated by the symbol — 1, written as an exponent. Thus 
the expression sin. -1 # denotes the arc whose sine is x. Hence, 
if y = sm.~ l x, we have sin. y—x. So also cos. -1 #, tang. _1 a;, etc., 
denote the arc whose cosine is #, the arc whose tangent is #, 
etc. Inverse trigonometrical functions are sometimes called 
circular functions. 

53. Differentiation of inverse trigonometrical functions. 
Formulas for the differentiation of inverse trigonometrical func- 
tions may be deduced from the formulas of Arts. 44-49. 

1. To differentiate y = sin. -1 #. 

Here the direct function is sin. y — x. Hence d sin. y—dx; 

dx 

that is, cos. ydy—dx, or dy — . 

^ u ? * cos. y 

But since sin. y=x, cos. y— Vl—x 2 . 

dx 
Hence dy=—= , radius being unity. 

Vl — x 2 

2. To differentiate y=cosr l x. 

Here cos. y—x. 

Hence —§m.ydy = dx, 

_ — dx —dx 

or ay—- — — - 

sin.?/ yi_aa 

3. To differentiate 3/ = vers. ~ l x. 
Here vers. y—x. 
Hence sin. ydy — dx. 



But sin. y = Vl — cos. 2 ?/ = Vl — (1 — vers, yf 

— Vl — (1— x) 2 = V2x—x 2 . 

Hence dy=z—= 

V'2x-x 2 

4. To differentiate 3/= tang r l x. 

Here tang. y—x. 

Hence &ec. 2 ydy=dx, 

dx dx 
or dy- 



"sec. 2 y 1 + x 2 ' 



62 DIFFERENTIAL CALCULUS. 

5. To differentiate y=cot r x x. 
Here cot. y—x. 
Hence — co$ec. 2 ydy = dx ; 

— — — CLX — ccx 

y ~~ cosec. 2 y ~~ 1 + x 2 ' 

6. To differentiate y=sec r l x. 
Here sec. y—x; 

tang, y sec. ydy—dx; 

r dx dx 

dy-- 



"tang. y sec. y~~xi/x 2 —l' 
7. To differentiate y^cosec." 1 ^. 
Here cosec. y=x; 

— cot. y cosec. ydy=dx; 

^ ~~ cot. 2/ cosec. y "~ a? -\/x 2 — 1 * 
Ex. 1. Differentiate the function y=sin . _1 



x 



Vl+x 2 



X 

Here sin. y 



Vl + x 2 

(lop 

Differentiating, we find cos. ydy~ § 

_ . 1 dx 

Hence dy= x o. 

cos.y (1+a;2) # 



/ X 2 ]_ 

But cos. y= VI— sin.2y=y 1-— — -j = r 

X +^ (l +a .2)7 

Hence, by substitution, 

i 6?a? dx 



1— « 2 
Ex. 2. Differentiate the function y=Bin.~ 1 1 2 . 

. _ 2cfe 

An*.dy=- T ^ ? . 

Ex. 3. Differentiate the function y=cos.~\4:X 3 — 3x). 



INVERSE TRIGONOMETRICAL FUNCTIONS. 63 

Here cos. y = 4# 3 — 3x. 

But by Trig., Art. 79, cos. 3&=4 cos. 3 £— 3 cos. b, 

or cos. y = 4 cos. 3 ( ^ )— 3 cos.j ^ 



Hence a?=cos.(^ 



—3dx 
Ans. dy- 



Ex. 4. Differentiate the function y=$mr\3x— 4# 3 ). 

. _ 3db 



Vl—x 2 

2x 
Ex. 5. Differentiate the function 2/= tang. -1 :: -. 

.A^s. dy—-T~ x — x. 

* 1 + x 2 

Ex. 6. Differentiate the function y=sin r 1 (2x— 1). 

Ans. ay—— ;==, 

Ex. 7. Differentiate the function 2/=sin. _1 m^. 

. , mdx 

Ans. dy=-y= 



- m c x 



2^2 



SB 



Ex. 8. Differentiate the function y— tang. ^ 

2<fe 

Ex. 9. Differentiate the function v=cos.~' , 

Ans. *y=Y+£r 

X 

Ex. 10. Differentiate the function y—tang." 1 

y 1— x 2 

Ans. dy=—i= 



x* 



64 DIFFERENTIAL CALCULUS. 



CHAPTEE III. 

SUCCESSIVE DIFFERENTIATION. EXPANSION OF FUNCTIONS IN SE- 
RIES. FUNCTIONS OF SEVERAL INDEPENDENT VARIABLES. VAN- 
ISHING FRACTIONS. 

54, Second differential coefficient. The differential coeffi- 
cient derived from any function of a variable is either a con- 
stant quantity or a new function of the same variable. This 
function may therefore be differentiated in the same manner 
as the original function ; and the differential coefficient of this 
second function is called the second differential coefficient of 
the original function. 

Thus, taking the function u = ax*, we have for the first dif- 
ferential coefficient, du 

-j- = 4:ax 3 . 
ax 

This coefficient is a function of x different from the original 
function. If we differentiate each member of the last equa- 
tion,wehave d ($\ =12a ^dx. 



fdu\ 
\dxj~ 



But since dx is supposed to be a constant, we have 

_, (du\ d(du) d.du . . . . . d 2 u 

d \7fi )=~fa~ = ~d^~> whlch 1S wntten ~d^' 

The symbol d 2 u denotes the differential of the differential of 

u, and is read second differential of u. 

Hence, we have d 2 u _^ m . 

—t- = 12ax 2 dx ; 

or, dividing each side by dx, 

-j-z=12ax 2 . 

where dx 2 denotes the square of dx, and is the same as if writ- 
ten (dx) 2 . 



SUCCESSIVE DIFFERENTIATION. 65 

The expression 12ax 2 being the differential coefficient of the 
first differential coefficient, is called the second differential co- 
efficient. 

Since the second differential coefficient is a function of x, 
we may differentiate it as a new function, and we obtain 

— — = 2iaxdx ; 
dx 2 

or, dividing each side by dx, we have 

dx 3 
which is the differential coefficient of the second differential 
coefficient, and is called the third differential coefficient. 

d 3 u 

The third differential coefficient -j-j is read third differen- 
tial of u divided by the cube of the differential of x. 

In the same way the fourth differential coefficient may be 
derived from the third, and so on. 

The symbol d 2 u must be carefully distinguished from du 2 , 
which denotes the second power of du ; and also from d(u 2 ), 
which denotes the differential of the second power of u. 

EXAMPLES. 

Find the successive differential coefficients of the following 
functions : 
1. u—ax n . 

Ans. —- = nax n ~ 1 ; - — =n(n—T)ax n ~ <i ; 
dx ' dx 2 K J ' 

d 3 u 

-— = n(7i—l)(n—2)ax n ~*; etc. 

(JLX 

It will be seen that by each differentiation the exponent of 
the power is diminished by unity. Hence, when n is a posi- 
tive whole number, the exponent will finally be reduced to 
zero, and the corresponding differential coefficient will be a 
constant; the next differential coefficient will therefore be zero, 
and the number of successive differential coefficients will be 
finite. But if the exponent be fractional or negative, the re- 



66 DIFFERENTIAL CALCULUS. 

suit can never become zero by the continued subtraction of 
unity ; hence the successive differential coefficients will always 
contain x, and the number of differential coefficients will be 
infinite. 

2. u=ax 3 +bx 2 . 

Ans. — = 3ax 2 +2bx: —-^ — Qax+2b:- T — — Ga. 
ax ax 2 ax 3 

3. u=x*+x 3 +x 2 +x+l. 

Ans.^=4:X 3 + 3x 2 +2x+l; ^i=,12x 2 +6x+2; 
ax ax 2 

4. w = 6# 4 — 4cx 3 — 5x 2 . 

Ans. ^=24:x 3 -12x 2 -10x: ^- = 72x 2 -24:X--l(): 
ax ax 2 

__144*-24, g^-144. 

5. w=a? -2 . 
^ s .^=_2or 3 ; ??L=6flr*; ^ = -24ar 5 ; 



1 

6. u=ax^. 



dx ' d# 2 " ' dx [ 

d A u 
dtf 



120ar 6 - etc. 



A du , _ l 6?% , _ 3 <# 3 ^ „ 5 



da? ' <fa> 2 ' dx 

7. 'w^sin. #. 



<#% , * 7 



. du d 2 u . <# 3 ^ 

^.^5. -7- = cos. ^ : -7—^-sin.a;; — — = — cos. x: 



dx ' dx 2 ' } dx 3 

d*u 



8. u=\og.(x + l). 



dx 



sin. x ; etc. 



^ = -6(^+1)-; etc. 



SUCCESSIVE DIFFERENTIATION. 67 

9. u=a*. 

Ans.-£=ar\og.a; 7 ^ r =a%\og.df; etc.; 



dx ° ' dx 2 



d"u ,,, . 
^— =a*(log.a) n . 



10. u=. 



1+x 2 



. du -2x dht 6x 2 -2 d 3 u 24^(1 -x 2 ) 
' dx (1 + x 2 f ' dx 2 (1 + x ? f ' dx 3 (1 + x 2 y ' ' 

55. There are four leading applications of the differential 
calculus in which integral calculus does not enter, viz.: 1. Ex- 
pansion of functions in series ; 2. Limiting values of functions 
which assume an indeterminate form ; 3. Maxima and minima 
values of functions ; and 4. Theory of curve lines and curve 
surfaces. We shall briefly consider each of these applications. 

maclattrin's theorem. 

56. The object of Maclaurin's theorem is to explain the 
method of expanding a function of a single variable into a se- 
ries of terms arranged according to the ascending powers of 
that variable with constant coefficients. 

Let u represent any function of x ; and if the function can 
be expanded in a series of positive integral powers of x, let 

u = A + Bx + Cx 2 + T)x 3 + Ex* + etc., (1) 

in which A, B, C, etc., are independent of x, and depend upon 
the constants which enter into the given function. It is re- 
quired to determine such values for the constants A, B, C, 
etc., as will cause the assumed development to be true for all 
values of x. 

Differentiating, we obtain the successive differential coeffi- 
cients of u, 

$=B + 2C£+3Dz 2 + 4Ez 3 + etc. (2) 

dx 

^=2C+2.3Dx+3.4Ex 2 + etc.; (3) 



68 DIFFERENTIAL CALCULUS. 

d 3 u 



dx 1 



- = 2.3D + 2.3.4Ez+ etc.; (4) 



etc., etc. 



But by hypothesis, the coefficients A, B, 0, etc., are indepen- 
dent of x; they must therefore remain the same whatever 
value may be ascribed to x; and if we can determine them for 
one value of x, we shall have their values in all cases. In 
equations 1, 2, 3, 4, etc., let us make x=0 ; and let (u) denote 

what u becomes under this hypothesis ; let J — I denote what 

du , (d 2 u\ , , d 2 u , , 

-=— becomes ; ( -— — denote what -^—- becomes, and so on ; 

dx ' \dx 2 J dx 2 ' 

the preceding equations furnish us 

(u) — A, therefore A — (u); 

(-)=,,B, - B^(-). 

Substituting these values of A, B, C, etc., in equation (1), it 
becomes 

, . (du\x (d 2 u\ x 2 (d 3 u\ x 3 

w= ^ + (^ji + (^jr2 + WJr^3 +etc -' ^ 

which is known by the name of Maclaurinh theorem. 

It will thus be seen that Maclaurin's theorem develops a 

function of one variable into a series of terms of w^hich the 

first is the value of the given function when x = ; the second 

term is x multiplied by the value of the first differential co- 

x 2 
efficient when x — ; the third term is ^ multiplied by the 

value of the second differential coefficient when x=0; etc. 
Ex. 1. Expand u=(a + x) 6 by Maclaurin's theorem. 
Differentiating, we have 



maclaurin's theorem. 69 

fdu\ 

6(a + x) 5 , hence I -7- I = 6a 5 ; 

g=5.6(a + *)', « (g)=5.6*; 

= 4 . 5 . 6 (a + x)\ « r -r-jf) = 4 . 5 . 6a 3 ; 

= 3.4.5. 6(a + x) 2 , " h^J=3.4.5.6a 2 ; 

=2.3.4.5.6(a+*), " ^^=2. 3. 4.5. 6a; 

= 1.2.3.4.5.6, " (SW.2.3.4.5.6. 

Here the differentiation terminates. Substituting these values 
in Maclaurin's theorem, we obtain 

/ \fi it . /> * . 5.6 A 9 , 4.5.6 , , , 3.4.5.6 , , 
u=(a + x) 6 = a 6 + 6a 5 x + ^— a 4 a 2 + a 3 x 3 + a 2 x 4 

. 2.3.4.5.6 s , 1.2.3.4.5.6x 6 

ax 5 + - 



du 
dx 
<Pu 

dx 2 
cPu 
dx 3 
dSi 
dx* 
d?u 
dx 5 
d 6 u 



T 2.3.4.5 T 2.3.4.5.6 ' 

or, reducing, we have z* = (a + x) 6 — a 6 + 6a 5 x + lSa 4 ^ 2 + 20a 3 £ 3 + 
15a 2 x 4 + 6ax 5 + x 6 , the same result as would be obtained by 
actual multiplication, or by the binomiaMheorem. 
Ex. 2. Expand u — {a+x) n by Maclaurin's theorem. 

— = n(a + x) n ~\ hence ( — - )= na n ~ l ; 

dx \dxj 

g=„(n-l)(a+ a! )-» J « (0} = n(n-l)a"-*; 

etc., etc. 

Substituting these values in Maclaurin's theorem, we have 

n 1 n(n — 1) _ _ 
u or (a + x) n — a n + - a""^ + a w ~ 2 x 2 + etc., 

1 1 . 2t 

which is the binomial theorem. 

Ex. 3. Expand u — —(a + x)~ l . 

a + x 

Ans. u-=ar l — a~ 2 x+a~ 3 x 2 —a~*x 3 + etc. 



70 DIFFERENTIAL CALCULUS. 

Ex. 4. Expand u — (a— x)~ 2 . 

A?is. ii = a~ 2 + 2cr 3 x + 3a~ i x 2 +4:cc 5 x 3 + 5ar 6 x i + etc. 
Ex. 5. Expand u = (a + x)~ 3 . 

Ans. u = a- 3 — 3a-*x + 6a- 5 x 2 — 10a- 6 x 3 + 15a _7 x 4 — etc. 

Ex. 6. Expand u=:(a+x) 2 . 

1 1 * 1 _3 1.3 _5 m 

Ans. u = a* + -a ^x———a 1 *x 2 +———a ^x 3 — etc. 
2 2.4 2.4.6 



Ex.7. Expand u=(a 2 +bx)*. 
Ans. u—a + 

Ex. 8. Expand u=(l+x 2 )^. 



A , a -1 for a~ 3 b 2 x 2 ar 5 b 3 x 3 
Ans. u-a-\ — — — + — jg etc. 



J 5 5x 8 
Ans. u =l + - -- + --— + etc. 

Ex. 9. Expand w=sin. x. 

Ans. u—x— + — + etc. 

2.3^2.3.4.5 2.3.4.5.6.7 



Ex. 10. Expand ^=cos. x. 
Ans. u — 1- 
Ex. 11. Expand u=\og. (1+x). 



Ans. - = 1-2+^1-2.3.4.5.6+ ^ 



rp~& /-y»4 



Ans. w=M(x— — +— — — + etc.). 
2 3 4 J 

Ex. 12. Expand u—a x . 

The successive differential coefficients have already been 
given in Ex. 9, Art. 54. For convenience, we will represent 
log. a by k. We shall then have 

, x . (dii\ 7 fd 2 u\ 7 (d 3 u\ „ 

Hence, by substitution, 

or=l+kx+ — -f^-^+etc. 

which is the exponential theorem. 
If a = e, the Naperian base, we have 

a r =,=l +x+ — +r ^ +r — + etc. 









MACLAURIs's THEOEEM. 71 

If x = l, we have 

c=1+1+ + 0^ + oW etc ' 

a series which gives the value of e the base of the Xaperian 
logarithms. 

57. ^l/»c in terms of its tangent. By means of Maclauiin's 

theorem, we can develop the arc in terms of its sine, tangent, 
etc., and hence obtain the value of the circumference of a 
circle of any given radius. 

Let ii — tang. -1 x. Hence (u) = 0. 

By Art. 53, (4), — = , which, by division, becomes 

\XX -L - r" X 

1— x 2 +x*— x 6 + etc. Hence (j-) = l. 

By successive differentiation, we have 

_ = _ 2x + 4x 3 - 6x 5 + etc., hence [ -— ) = ; 
dx l \dx 2 / 

g' = - 2+ 3.^-5.6*. + e,c, " (S) = - 2 > 
g =2 .3.4.-4.5.6*» + etc., - (£)=»; 

^=2.3.4-3.4.5.6.* + etc., - ©=2-3.4; 

etc, etc. 

Substituting these values in Maclaurin's theorem, we have 

2a; 3 , 2.3.4a; 5 

or i»=tang." 4 *=-— — +-— etc., (1) 

which is an expression for the length of an arc in terms of its 
tangent. From this development we are enabled to find the 

7T 

value of 7r. Thus, let w=45°= j, whence x = l, and we have 
arc45° = l-KW + *-Tr + :nr- etc. 



72 DIFFERENTIAL CALCULUS. 

From this series an approximation may be made to the cir- 
cumference of a circle, but the series converges so slowly that 
an accurate result can not be obtained from it without very 
great labor. 

58. Computation of the value of ir. In order that series 
(1), Art. 57, may converge rapidly, x must be a small fraction. 
This object may be accomplished by employing the principle 
that the arc 45° is equal to the arc whose tangent is ^ plus 
the arc whose tangent is ^. 

This principle may be proved as follows. From Trig., 

Art. 77, ■ tang, a + tang, b 

tang. (a+b)=z i — f — r* , . 

° v ' 1 — tang, a tang, b 

Assume a + b = 45°, and tang. a=£. 

Then tang. 45° = 1 = i+taiig.fc w i ience tang< 6 _ i 

& 1— J tang. V 6 

Hence 45° = the arc whose tangent is £+ the arc whose tan- 
gent is £. 

Substituting -| and £ for as in equation (1), Art. 57, we have 

45 = I = 2-^23+ 5 -^i- r 27+ etc 

+ 1-J-+J-— i-+etc 

In order to compute the value of this, or any similar series, 
reduce each term to a decimal fraction. A few terms of this 
series five it 

8 1=0.4636476 + 0.3217506 = 0.785398. 

Hence tt = 3.141592 + . 

59. When Maclaurirfs theorem fails. Maclaurin's theorem 
enables us to develop any function of one variable, whenever 
the function can be developed in the proposed form. But 
there are functions which can not be expanded in a series of 
positive integral powers of the variable. In such a case the 
function, or some one of its successive differential coefficients, 
becomes qo when the variable is made equal to 0. 



maclaurln's THEOREM. I'd 



For example, let u=x*. Hence (w)= 0. 

du_ 1 

dx 2 J 

d 2 u 1 



(du\ 1 

Substituting these values in Maclaurin's theorem, we have 






-I a; 2 

u — x' z = + oox— x . — + etc. ? 

a series of which we can make no use. In this example, Mac- 
laurin's theorem is said to fail, and it fails because the f unc- 

i 
tion x 1 can not be expanded in the particular form assumed 

in the theorem. 

So also the functions u — -^ K = cot. x, ^=log. x, can not be 

x 

developed by Maclaurin's theorem in integral powers of x. 
Such functions can, however, often be developed in fractional 
powers of x, or in powers of some simple function of x ; as, 
for instance, in powers of 1 + x. 

60. Function of the sum of two variables. A quantity is 
said to be a function of the sum of two variables when in the 
algebraic expression for the function a single variable may be 
substituted for the sum, and the original function be thus re- 
duced to a function of the single variable of the same form 
as the primitive function. 

Thus, let u = a(x + y) n . 

If in place of x+y we substitute z, the function becomes 
u — az 1 ^ which is a function of z, of the same form as a{x+y) n . 

So also the function u=\og. (x—y) becomes u=}og. z when 
we put z for x—y. Here u is a function of the sum of x 
and — y. 

61. Partial differential coefficients. Let u be a function 
of two independent variables x and y ; then, since the varia- 
bles are independent, however either be supposed to vary, the 

D 



74 DIFFERENTIAL CALCULUS. 

other will remain unchanged, the function .ought to furnish 

two differential coefficients ; the one resulting from ascribing 

a variation to a?, and the other from ascribing a variation to y ; 

y entering the first coefficient as if it were a constant, and x 

entering the second as if it were a constant. 

If we suppose y to remain constant and x to vary, the differ- 

cLix, 
ential coefficient will be -j- ; and if we suppose x to remain 

constant and y to vary, the differential coefficient will be -r-. 

The differential coefficients which are obtained under these 
suppositions are called partial differential coefficients. The 
first is the partial differential coefficient taken with respect to 
Xj and the second is the partial differential coefficient taken 
with respect to y. 

taylor's theorem. 

62. Taylor's theorem is a theorem for expanding a function 
of the sum of two variables into a series arranged according 
to the ascending powers of one of the variables, with coeffi- 
cients that are functions of the other variable, and depend 
also upon the constants which enter the given function. 

The demonstration of Taylor's theorem is founded upon the 
principle that if u'=f{x + y) be any function of x + y, the par- 
tial differential coefficient of u will be the same, whether we 
suppose x to remain constant and y to vary, or y to remain 
constant and x to vary ; that is, the partial differential coeffi- 

du -. du , 

cients -=- and — — are equal. 
ax ay 

Assume z — x + y, then %i—f{z). 

If we differentiate this as a function of x, y being regarded 

as a constant, we shall have, Art. 33, 

du _du dz 

dx dz dx 

If we differentiate it as a function of y, x being regarded as 

a constant, we shall have 



75 

du du dz 

dy~ dz dy' 
But, since 2 =#+?/, when y is regarded as constant, dz — dx; 
that is, ^^ 

CLJC -j 

r Also, when x is regarded as constant, dz—dy; that is, — =1. 

c/ 
^ ^ da du 

Hence -— — -— and — =— ; 

du du 

dx dy } 
that is, the partial differential coefficient of the sum of two 
variables, x and y, obtained on the supposition that x is va- 
riable and y constant, is the same as that obtained on the sup- 
position that y is variable and x constant. 

Ex. 1. Find the two partial differential coefficients of 

u = (x+y) n . du . v , _ du v , 

Jl Ans. -^=n(x + yy-\ and ^=n(x+y)*~ l . 

Ex. 2. Find the two partial differential coefficients of 
i 
u—(x + y)^. . du .. 2 .du ly _a 

J) Ans. ^=*{x+y) * and -^=^(x + y) * 

Ex. 3. Find the two partial differential coefficients of 

^ = loo\ (x + y). . du 1 . du 1 

Ans. -j~ = — ; — , and -j-= — ; — . 
ax x + y' ay x + y 

63. Product or quotient of two variable quantities. The 
preceding principle will not hold true for the partial differen- 
tial coefficients of the product or quotient of two variable 
quantities. 

Ex. 1. Find the two partial differential coefficients of 
(x\ n du nx n ~ l _ du —nx n 

\y/ dx y n ' dy y v + l 

Ex. 2. Find the two partial differential coefficients of 

u = \og.(xy). du 1 du 1 

Ans. -j-=- and -j-=-. 
dx ar rfy y 



76 DIFFERENTIAL CALCULUS. 

64. Demonstration of Taylors theorem. Let v! be any 
function of the sum of two variables x and y. It is required 
to determine the law of the development when the function 
can be expanded in the form 

u' =f(x +y)=A+By + Cy* + Vy* + etc, (1) 
where A, B, C, D, etc., are independent of ?/, but dependent 
upon x, and upon the constants which enter the primitive 
function. It is required to find such values for A, B, C, etc., 
as shall render the development true for all possible values of 
x and y. 

If we differentiate upon the supposition that x varies and y 
remains constant, and divide the result by dx, we shall have 

du' _dA dB dC 2 dT> 3 

dx dx dx dx dx 
If we differentiate upon the supposition that y varies and x re- 
mains constant, and divide by dy, we shall have 

'-p=B + 2Q/+3D?/ 2 +etc. 

But, by Art. 62. we have -— — -—; therefore 

' J dx dy 

dA dB dC _ ^ _ ~ OT . 

d^ + d^ V+ dx y + etc.=B + 2Cy + 3D?/ 2 + etc. 

Now since this equality exists whatever be the value of y, 
the coefficients of the like powers of y in the tw r o members 
must be equal to each other (Algebra, Art. 378) ; therefore, 



— =B 

dx 


(2) 


§-*>■ 


(3) 


^ = 3D, etc. 
dx 


(*) 



But, since equation (1) is true for all values of ?/, we may 
make y=0, in which case function of x + y will reduce to 
function of #, which we will denote by u. Therefore, A = u. 



taylor's theorem. 



Substituting this value of A in equation (2), we have 

du 

~ dx' 

Substituting this value of B in equation (3), we have 

^^ d.du d 2 u . ^ d 2 u 

2(3=^ r"="TT5 whence Cm 7 

dx . dx dx 1 ' zdx z 

Substituting this value of C in equation (4), we have 
rtX ^ d.d 2 u d 3 u . _ d 3 u 

3D= 23£^=2^ wh6DCe D = 2T3^' and S ° ° D - 

Substituting these values of A, B, C, D, etc., in equation (1), 

which is known by the name of Taylor's theorem. 

It will thus be seen that Taylor's theorem develops a func- 
tion of the sum of x and y into a series of terms that contain 
increasing powers of y multiplied by coefficients which are 
functions of x. The first term is the value of the original 
function when y=--0; the second term is y multiplied by the 
first differential coefficient of the first term with respect to x ; 

y 2 
the third term is ^- multiplied by the second differential co- 
efficient of the first term with respect to x ; etc. 

The development oif{x—y) is obtained from the same the- 
orem by changing + y into — y, thus, 

, -, x die y d 2 u y 2 d 3 u y 3 

Ex. 1. Expand u' = {x+y) n by Taylor's theorem. 
Making y=0, we obtain u=x n , and thence, by differentiation, 
du . d 2 u , WN , 

~=n(n-l)(n-2)x«-\ etc. 

These values being substituted in the formula, give 

, , , n(n—T) „ n n(n—l)(n—2) , 

u'=(x+y) n =x n +nx n - 1 y+ ' 2 V ~ 2 y 2 + 123 y+ etc -> 

the same as found by the binomial theorem. 



78 DIFFEEENTIAL CALCULUS. 

Ex. 2. Expand u' = (x—y)~ 4c by Taylor's theorem. 

Ans. (x-y)- 4 =x~*+4:x- 5 y+10x- 6 y 2 +20x- 1 y 3 +35x- 8 y 4: + etc. 

i 
Ex. 3. Expand u f =z(x—yY by Taylor's theorem. 

i i _2 2_5 2 . 5 _ 8 

An8.(x—y)^=x^—^x ^y—o~a x ^y 2 — o a q x ^2/ 3 — etc. 

Ex.4. Expand 'w'^sin. (x+y) by Taylor's theorem. 
Ans. sin. (*+y)= S in. ® j ^-^2+ lm2 .8.4Tl.2.8A.6.e +eto - ] 

( y3 yb yl | 

+ cos. x j 2/-J7273 + 1.2.3.4.5""1.2.3.4.5.6.7 + etC ' j 
Ex. 5. Expand ^'^cos. («+y) by Taylor's theorem. 

( y 2 y* y 6 1 

Ans. cos. (x+y)=cos.x < 1 "X72 + 1.2.3.4 ~ 1.2.3.4:.5.6 + etc * J 

j y 3 y 5 y 1 [ 

- sm ^|2/~l7273 + i.2.3.4.5""l .2.3.4 .5.6.7 + etc ' j 
Ex. 6. Expand u'=log. (x + y) by Taylor's theorem. 

-4w.log.(«+y)=log.flJ+|-^+^-£i+eto. 
Ex. 7. Expand w' = # x + y by Taylor's theorem. 

( y 2 y 3 ) 

J.715. a x + v =a* 1 1+y . log. &+T— o(log. ^) 2 +T-^— q(log. a) 3 +etc. > 

65. TFA<m Taylor's theorem fails. Taylor's theorem will 
give the expansion of (x + y) in all cases so long as x retains 
its general value ; but in certain functions a particular value 
may be given to x which will render some of the differential 
coefficients infinite ; in such cases the theorem is said to fail 
in giving the development. In the demonstration of the the- 
orem, Art. 64, when the coefficients of the like powers of y in 
the two members of the equation were assumed to be equal, 
these coefficients were supposed to be finite quantities. Hence, 
if such a value should be assigned to one of the variables as 
to make any one of these coefficients indefinitely great, the 
demonstration would be inconclusive. In such a case the the- 
orem fails in giving the development; and this failure indi- 



taylor's theorem. 79 

cates that the particular value of the function proposed can 

not be expressed in integral positive powers of y combined 

withjinite coefficients. 

i 
Ex.1. Develop u / = (x + y + ay r by Taylor's theorem. 

i 
Here •u = (x + ay 2 

ax 2(x + ay 

d 2 u , -3. 1 



-=-#x + a) 2 : 



d 3 u 9/ 5 3 

-r- 3 = *(x+a) *= H ; etc. 

Ux S{x + af 

Substituting in Taylor's theorem, we have 

JL i y y 2 3?/ 3 
u'=(x+y+a) 2 =(x+a)*+ — - — j- — - — 3+ — g-retc, 

2(x+ay 8(x+ay ±8(x+a)* 
which is the general development of u f . But if we suppose 
x= — a, the series becomes 



^ r = ?/ T — O + oo — oo + oo— etc. 



from which nothing can be determined except that the square 
root of y can not be exhibited in integral positive powers of y 
with finite coefficients. 

Ex. 2. For what value of x does the development of %l' — 

i 

(x + y+by by Taylor's theorem fail ? Ans. x——b. 

DIFFERENTIATION OF FUNCTIONS OF TWO INDEPENDENT VARIABLES. 

66. Let u be a function of x and y, where each variable is 
supposed to be entirely independent of the other. We will 
suppose the variables to change separately, x first receiving an 
increment A, and y afterward receiving an increment k. 

When x receives the increment h,f(x, y) becomes y(a?+ A, y) ; 
and if y be regarded as constant, this function may be devel- 
oped by Taylor's theorem ; whence 

, „ _ . clu. d 2 u 7i 2 , H . 

u =f(x + h,y) = u + ^h+-^ j+ etc., (1) 



80 DIFFERENTIAL CALCULUS. 

in which y~,"3~^> etc., are the differential coefficients of u 

taken under the supposition that x alone is variable, and they 
are all functions of x and y. 

If in equation (1) we suppose y to become y + k, each term 
will receive a change. The first member becomes/^ + A, y + Jc). 
The first term u of the second member may be expanded by 
Taylor's theorem, and becomes 

du 7 dhc k 2 
U +dy k + W 2 + etC> 
In like manner — in the second term of equation (1) may 

(JuX 

be expanded, and becomes 

7 du 

du dx _ 

di + ~d^ k+ etc -' 

du d 2 a _ 
or T"+ 7 7 # + etc. 

aa? dxdy 

So also -T-2 in the third term may be expanded into a series 

/72/iy 

of terms, of which -j-^ is the first, and is the only one which 

does not contain h as a factor. 

Making these substitutions in equation (1), we have 

„ n , 7 7N du y d 2 u k 2 

u'=f(x+h,y + k) = u + -^k + -^r ^"+ etc., 

du d 2 u 77 

+ d^ h+ d^h/ k+ etc -' 

d 2 u h 2 

where cacli of the omitted terms involves more than two di- 
mensions of It and k. 

If from each member of equation (2) we subtract f(x, y), 

we have u" — u — ~j-h-\--j-k + other terms involving at least 

two dimensions of h and k. This is the expression for the 
increment which the function u has received from attribu- 



FUNCTIONS OF SEVERAL VARIABLES. 81 

ting an increment to each of the variables x and y ; that is, 

du . die . , 

mcr. u — -t- % incr. #+-r~ mcr. y+ terms involving two dimen- 
sions of A and k. 

Dividing each member by incr. u, we have 

du . incr. x du . incr. y _ 

/Yn mm 1 ^/ ' 



dx . incr. ^ <:&/ . incr. ^ 
where X represents the omitted terms, each of which contains 
as a factor h or k in at least two dimensions, and also contains 
increment t^asa divisor. 

Kow, to determine the value of these ratios at their limit, we 
must make h and k each equal to zero, in which case the term 
X disappears, as in Art. 18, and we have 

dudx dudy 
~ dxDu dyDu? 
_ dudx dudy 

where Du in the first member denotes what is called the total 
differential of the function, while the first du in the second 
member denotes what is called the partial differential taken 
with respect to x, and the second du denotes the partial dif- 
ferential taken with respect to y. 

The total differential T)u is by some authors written with 
the small letter, as du, in which case the partial differential is 

, n r du du . 

made the numerator of a traction, as -4- or -y-, which ar£ par- 
tial differential coefficients of u with respect to x and to y. 

67. Total differential. The differential of a function of 
two independent variables, which is obtained upon the suppo- 
sition that both the variables have changed their values, is 
called the total differential of the function ; and we see from 
Art. 66 that the total differential of a function of two inde- 
pendent variables is equal to the sum of the partial differ- 
entials. 

D 2 



82 DIFFERENTIAL CALCULUS.. 

In the same manner it may be proved that the total differ- 
ential of a function of any number of variables is the sum 
of the several partial differentials arising from differentia- 
ting the function relatively to each variable in succession as 
if all the others were constants. 

Ex. 1. If one side of a rectangle increase at the rate of 1 inch 
per second, and the other at the rate of 2 inches, at what rate 
is the area increasing when the first side becomes 8 inches, and 
the last 12 inches ? Ans. 28 inches per second. 

Ex. 2. If one side of a rectangle increase at the rate of 2 
inches per second, and the other diminish at the rate of 3 inch- 
es per second, at what rate is the area increasing or diminish- 
ing when the first side becomes 10 inches, and the second 8 ? 

Ans. 

Ex. 3. If the major axis of an ellipse increase uniformly at 
the rate of 2 inches per second, and the minor axis at the rate 
of 3 inches, at what rate is the area increasing when the major 
axis becomes 20 inches, the minor axis at the same instant be- 
ing 12 inches ? Ans. 21tt inches. 

Ex. 4. If the altitude of a cone diminish at the rate of 3 
inches per second, and the diameter of the base increase at the 
rate of 1 inch per second, at what rate does the solidity vary 
when the altitude becomes 18 inches, the diameter of the base 
at the same instant being 10 inches? Ans. 

Ex. 5. Find the total differential of u=x 3 + ax 2 y + y 3 . 

dudx /ft o _ . _ 
— — — = [3x 2 + 2axy)dx, 

^=(ax 2 + Sf)dy. 

Hence T>u = (3« 2 + 2asey)dx + (ax 2 + 3y 2 )dy. 

x A~ ?/ 
Ex. 6. Find the total differential of u- 



x-y 

_ 2xd?j—2ydx 
Ans. Du= — -r- — ~ — . 
(x-yf 



DIFFERENTIAL EQUATIONS. 83 

Ex. 7. Find the total differential of u = - + -. 

V x 

_ dx ydx dy xd?/ 
Arts. D^=— — ^r+ — —- -#. 

y x 2 x y l 

Ex. 8. Find the total differential of ii = ax 2 y 3 z*. 

Aiis. Du = 2axy 3 z i dx + 3ax 2 y 2 z*dy + ±ax 2 y 3 z 3 dz. 

Ex. 9. Find the total differential of u—x v . 

Am. Du=yx y ~ 1 dx+x y log. xdy. 

x 
Ex. 10. Find the total differential of w = tang . -1 -. 

£> y 

^ ydx— xdy 
Am. Du = 9 , 9 . 

Ex. 11. Find the total differential of u = $m.(xy). 

Am. T>u = cos. (#y) { ?/<f # + a%?y { . 

DIFFERENTIAL EQUATIONS. 

68. If we differentiate the equation of a line, we shall obtain 
a new equation; which expresses the relation between the dif- 
ferentials of the co-ordinates of the line. This equation is 
called the differential equation of the line. 

If we take the equation of a straight line 

y=mx+c, (1) 

and differentiate it, we obtain 

a result that is the same for all values of c, and is equally ap- 
plicable to every line that is parallel to y=mx. 
Differentiating equation (2), we obtain 

d 2 y „ 
— -= 

dx 2 

This last equation is entirely independent of the values of 
the constants m and c, and is equally applicable to every 
straight line which can be drawn in the plane of the co-ordi- 
nate axes. It is called the general differential equation of 
lines of the first order. 



84 DIFFERENTIAL CALCULUS. 



69. If we take the equation of the circle 




y 2 =r 2 — x 2 , 


(1) 


and differentiate it, we obtain 




2ydy——2xdx; 




. dy x 
whence —-=—-. 

dx y 


(2) 


Equation (2) is independent of the value of the radius r, and 


hence it belongs equally to every circle having its 


origin at the 


centre. 




If we take the equation of the parabola 




y 2 = 4:ax, 


(i) 


and differentiate it, we obtain 




2ydy=4cadx; 




_ dy 2a 
whence -f-= — . 

dx y 


(2) 


But from equation (1) we have 




v 2 





hence equation (2) becomes 

dx~2x 
This equation is independent of the value of the latus rectum 
4&, and hence it belongs equally to every parabola whose prin- 
cipal vertex is at the origin, and whose axis is the axis of x. 

70. If we take the general equation of lines of the second 
order, which is (Anal. Geom., Art. 234) 

y 2 =zmx + nx 2 , (1) 

and differentiate it, we obtain 

2ydy = mdx + 2nxdx. (2) 

Differentiating again, regarding dx as constant, we obtain 

2dy 2 + 2yd*y=2?idx 2 , 
or, dividing by 2, dy 2 +yd 2 y=ndx 2 . (3) 

By combining equations (1), (2), and (3), we may eliminate the 
constants m and ?i, and we obtain 

y 2 dx 2 + x 2 dy 2 + yx 2 d 2 y -— 2xydxdy = 0, 



VANISHING FRACTIONS. 85 

which is the general differential equation of lines of the sec- 
ond order, and is applicable to every conic section of which 
the axis of x is a diameter and the origin is a principal vertex. 

71. Thus we see that an equation may be freed from its 
constants by successive differentiations; and for this purpose 
it is necessary to differentiate it as many times as there are 
constants to be eliminated. The differential equations thus 
obtained, together with the given equation, make one equation 
more than the number of constants to be eliminated, and hence 
we may derive from them a new equation, which will be freed 
from these constants. 

The equation which is thus obtained is the general differen- 
tial equation of the species of lines one of which is represented 
by the given equation, since, being independent of the con- 
stants, it belongs to all lines of the same species referred to 
the same co-ordinate axes. 

Generally it is not difficult, by differentiation and elimina- 
tion of constants, to deduce differential equations ; but to pass 
from differential equations to the primitive equation from 
which they were deduced is more difficult, and forms one of 
the most important branches of the mathematics. In fact, 
nearly all investigations in physical science lead to differential 
equations, from which it is required to discover the relation 
subsisting among the primitive variables. 

VANISHING FRACTIONS. 

72. It sometimes happens that the substitution of a particu- 
lar value for the variable in a fraction causes both the numer- 


ator and denominator to vanish, and the fraction reduces to ~. 

, . x— a 

Such fractions are called vanishing fractions. Thus — — — 2 

x a 

becomes ~ when x—a. 

This form arises from the existence of a factor common to 



86 DIFFERENTIAL CALCULUS. 

both numerator and denominator, which factor becomes 
under the particular supposition. Hence, if we can by any 
means eliminate this factor before substituting a for x, we 
shall be able to find the true value of the fraction. 

Sometimes the vanishing factor is readily seen, and may be 
immediately canceled ; as in the above fraction, which con- 
tains the factor x— a in both numerator and denominator. 

Canceling this factor, the fraction becomes — ■ — , whose value 

. X-\- (X> 

is 7T- when x = a. This is, therefore, the true value of the frac- 

2a ' ' 

x— a 
tion — - 9 when x — a. 

x 2 — a z 

So also the fraction 

x 2 —ax 

^z^=o when •=«/ 

but rejecting the common factor x—a, the fraction becomes 

x a 

t, which reduces to t when x~a. 

In many other cases the true value of a vanishing fraction 
may be ascertained by performing a few simple transforma- 
tions. Thus j_ 

(x*-a 2 )* + (x-a) 



(x-a) 2 + (x 2 -a 2 y 2 



r =r- when x — a; 



"0 



but dividing numerator and denominator by the common fac- 

tor {x— ay\ we have 

(x+a)* + (x-a)* V2a 

— i = ;= when x—a. 

\ + (x + af 1 + V'2a 

73. The differential calculus affords a more direct and gen- 
eral method of finding the value of a vanishing fraction. Let 

N 
the fraction be ^ ; then, if N and D both vanish w T hen a par- 
ticular value a is assigned to the variable, they must have a 
common divisor consisting of some power of x— a. 



VANISHING FRACTIONS. 87 

N" Y(x— a) 
First suppose t) = q, ,_ .y in which P and Q are functions 

of a?, but do not contain the factor x—a. Differentiating, we 
have dN dP(x-a) + Tdx 

dD~d Q(x —a) + Qdx' 
P 
which becomes ^r when we substitute a for x, and since P and 

P 

Q do not vanish when we substitute a for x, ^ is the required 

value of the fraction. 

Next suppose the common factor to be (x— a) 2 , or 

N P(x-a) 2 
V-Q(x-a) 2 ' 
dN dP(x-a) 2 + 2F(x-a)dx 
then dD ~ dQ(x - a) 2 + 2Q(aj - a)dx> 

which =jt when a is substituted for x. 

Differentiating a second time, we have 

d 2 ^ _ d 2 ¥(x-a) 2 + 4:dP(x— a)dx+ 2Pdx 2 
d 2 D~d 2 Q(x-a) 2 + 4:dQ(x-a)dx + 2Qdx 2 ' 

P 

which becomes tt when we substitute a for x, and since P and 

^ P . 

Q do not vanish when x—a^ ^ is the required value of the 

fraction. 

„ . r N T(x— a) m . . ... 

Generally, 11 ^ = t=w r^, where m is a positive integer, by 

differentiating N and D m times, and substituting a for x in 

1ST P 
the result, we shall obtain =f)=7Y 

mA T _ N F(x-a) m P. x _ ,__ 

74. It ^=7^7 ^r=7sr(^ — ^r - ", and we differentiate n 

D Q(x—a) Q v J ' 

times, and substitute a for a? in the result, we shall have 

N 

p: = Q=:0, m being greater than n ; but if m be less than n. 



88 DIFFERENTIAL CALCULUS. 

N 
differentiating m times we shall have t^ 00 - Hence, to find 

the value of a vanishing fraction, for the numerator and de- 
nominator substitute their first differential coefficients, their 
second differential coefficients, and so on, till we obtain a frac- 
tion in which numerator and denominator do not both vanish 
for x — a; the value thus found will be the true value of the 
vanishing fraction. If the numerator vanish when the de- 
nominator is finite, the fraction =0, but if the denominator 
vanish when the numerator is finite, the fraction = oo. 

The preceding demonstration is only applicable to algebraic 
functions ; but it may also be shown of transcendental functions, 

that when they take the form of ~ they have a determinate 

value, which may be found by successive differentiations. 

x 3 — 1 
Ex. 1. Find the value of u= o , n 9 k when x=l. 

x 3 + 2x 2 — x—2 

Here the differential coefficient of the numerator is 3x 2 , and 
that of the denominator is 3# 2 + 4#— 1. 

3x 2 1 , 

Ex. 2. Find the value of u — =- when x=l. 

X 1 A 

Ans. u — n. 

/£>* rA 

Ex. 3. Find the value of u—-^ when x~a. 

gQ* r/2 

Ans. u=2a 2 . 

qn$ /ytZ qn I "j 

Ex. 4. Find the value of %i— — i — ,, 3 2 when x~\. 

X jLiX -j- X 

Ans. u~2, from two differentiations. 
3^2 _ q x I 3 
Ex. 5. Find the value of u—-^-^ — a , 9 when x=l. 

Ans. w=t:. 
_. . _.. , . _ n ax 2 — 2acx+ac 2 . 
Ex. 6. Find the value of u =f x rZ2 bcx+bc 2 whe11 X=G ' 

Ans. u — -r. 
b 



VANISHING FRACTIONS. 89 

Q* Jf 

Ex. 7. Find the value of w= when x—0. 

Arts. i^ = log. ( j J. 

Ex. 8. Find the value of u=-j — -, — , , ,> q ; when x = a. 

ar — 'lax 6 + za 6 x — a 4 

Ans. u — ^o. 

^ sin x 

Ex. 9. Find the value of u= — when #=90°. 

cos. x 

Ans. u — 0. 

75. The method of Art. 74 is not applicable when the com- 
mon factor x—a of the numerator and denominator is a rad- 
ical. For, since at each differentiation the exponent m of the 
factor is diminished by unity, if the exponent is a proper frac- 
tion it will become negative without reducing to zero, and the 

fraction will take the form of — . 

00 

In such a case, we may obtain the true value of the fraction 
by substituting a + h for x both in the numerator and denom- 
inator of the fraction, reducing the result to its simplest form, 
and then making the increment equal to zero. It is plain that 
this will amount to the same as the substitution of a for x ; but 
by the reduction of the fraction to its lowest terms, the com- 
mon factor is made to disappear, and we obtain the true value 
of the fraction for the particular value of the variable. 

The substitution of a particular value for the variable may 
cause a function to assume the form of Ox oo, 0°, or go , etc. 

Such expressions can sometimes be reduced to the form ~, and 
their values be determined by the methods already explained. 

( nrp> /y2 \ 2 

Ex. 1. Find the value of u — j- when x=a. 

{x—cif 

Assume x = a + h; 



3 



then ^= v : ^— '—=(?a+hy, 



90 DIFFERENTIAL CALCULUS. 

3 

which, when A = 0, becomes (2a) 1 *, which is the value of the 
given fraction when x~a. 

2 

T3i ^ ti. j x i i o (x 2 —3ax + 2a 2 Y 

Ji,x. 2. Jbmd the value or u—- — — when x — a. 

(x 3 -ay 

Assume x=a+h; 

then 

N (h 2 -ahf hhh-af 

jv = 1 = j-= j when #=# ? 

^ (3a 2 h + 3ah 2 + A 3 ) T (3a 2 + 3a A + A 2 ) T (3a 2 f 

Or ^zz:^z=:0. 



1 J_ 

"/ 2 l / /» /v\ 2 



Ex. 3. Find the value of u~ j — — when x~a. 

(x 2 —a T f 

A i 

Ans. u 



(2af 



v 2o^X X^ v fl.T^ 

Ex. 4. Find the value of u— j= when x—a. 

a— V ax. 

Ans. it— ha. 



GEOMETRICAL REPRESENTATION OF A FUNCTION. 



91 



CHAPTER IV. 

GEOMETRICAL REPRESENTATION OF THE FIRST DIFFERENTIAL CO- 
EFFICIENT. MAXIMA AND MINIMA VALUES OF A FUNCTION. 

76. Geometrical representation of a function. Every func- 
tion of a single variable may be represented by the ordinate 
of a curve, of which the variable is the corresponding abscissa. 
For let y be an explicit function of a variable x expressed by 
an equation y—f{x)^ 

and let any value be assigned to x, and the corresponding value 
of y be deduced. Suppose x and y to be the co-ordinates of a 
point referred to two rectangular axes ; then for each value of 
x there will be one or more values of y, and these, if real, will 
determine the position of points with respect to the given axes. 
The points thus determined will lie upon and represent a locus 
or curve, and in general the relation between the ordinate and 
abscissa of any point of the curve will be the same as that be- 
tween the given function and the independent variable. 




77. First differential coefficient. Let CPP' be a curve 
whose equation is y=f(%), and let p, 

P be any point of this curve whose 
co-ordinates are x and y. Increase 
the abscissa CR or x by the arbi- 
trary increment RR', which we 
will represent by h. Denote the 
corresponding ordinate P'R' by y' ; draw the secant SPP', 
the tangent PT, and draw PD parallel to CX. Then 
P'D^P'R'-PR^^-^^AA + BA^Art^S). 
But from the triangle P'DP we have 

PD : PD : : 1 : tang. PTD=^, 



92 DIFFERENTIAL CALCULUS. 

and substituting for P'D and PD their values, we have 
tang. P'PD=tang. PSB= "^ =A + Bh, 

which expresses the ratio of the increment of y to that of x. 
In order to find the differential coefficient of y with respect to 
x, we must determine the limit of this ratio by making the 
increment equal to zero (Art. 18). 

Now, when h is diminished, the point P' approaches P, the 
secant PS approaches the tangent PT ; and finally, since the 
limiting position of the secant is the tangent, we have at the 
limit, making A = 5 

tang.PTX=A=||; 

that is, the tangent of the angle which a tangent line at any 
point of a curve makes with the axis of abscissas is equal to 
the first differential coefficient of the ordinate of the curve. 

78. Given values of the first differential coefficient. If it 
is required to find the point of a given curve at which the tan- 
gent line makes a given angle with the axis of X, we know 
that at this point the first differential coefficient must be equal 
to the tangent of the given angle. If we represent this tan- 
gent by a, we must have 

dy 

dx~ ' 
and this, combined with the equation of the curve, will give 
the particular values of x and y for the required point. 

If the tangent line is to be parallel to the axis of X, the 
angle which it forms with that axis is zero, and its tangent is 

zero ; hence, for the point of tangency ~^ = 0. 

If the tangent line is to be perpendicular to the axis of X, 

du 
then, since the tangent of 90°= oo, ~--=lqo. That is, for a point 

at which the tangent to a curve is parallel to the axis of ab- 
scissas, the first differential coefficient is equal to zero; and 



TANGENT TO A CURVE. 93 

for a point at which the tangent to a curve is perpendicular 
to the axis of abscissas, the first differential coefficient is eqiial 
to infinity. 

Ex. 1. It is required to find the point on a given parabola at 
which the tangent line makes an angle of 45° with the axis. 
The equation of the parabola (Anal. Geom., Art. 85) is 

y 2 = 4tax. 
Differentiating, we obtain 

2ydy=4:adx y 
dy 2a 
dx~ y ' 
But, since tang. 45° equals radius or unity, we have 

— = 1, or 2a— y. 

y ' J 

Combining this with the equation y 2 — 4:ax J we find 

x — a. 

Hence the required tangent passes through the extremity of 

the ordinate drawn from the focus. 

Ex. 2. It is required to determine at what point the tangent 

to the circumference of a circle is parallel to the axis of X. 

The equation of the circle (Anal. Geom., Art. 60) is 

x 2 + y 2 — r 2 . 

By differentiating we obtain 

dy x 

dx~ ~~ y' 

and placing this equal to we find 

x=0. 

But when x = we have y—^r; 

hence the tangent, is parallel to the axis of X at the two points 

where the circumference intersects the axis of ordinates. 

Ex. 3. It is required to determine at what point the tangent 

to an ellipse is perpendicular to the axis of X. 

The equation of the ellipse (Anal. Geom., Art. 121) is 

x 2 y 2 _ 
— A-- — 1 

~2 i 7.2 — - 1 - 



a 



ir 



94 



DIFFERENTIAL CALCULUS. 



By differentiating we obtain 

dy h 2 x 
dx~ a 2 y' 

and placing this 'equal to oo, we have 



~a 2 y' 



:00. 



whence y = 0, and x=±a. 

Hence the tangent is perpendicular to the axis of X at either 

extremity of the transverse axis. 

MAXIMA AND MINIMA OF FUNCTIONS OF A SINGLE VARIABLE. 

79. Definition of maximum and 7ninimum. Suppose u to 
be a certain function of x, and that while the variable x 
changes gradually from one definite value to another, u 
changes in such a manner that it is sometimes increasing and 
sometimes decreasing. There must, then, be a certain value 
of x, for which u, having previously been increasing, begins 
to decrease ; or begins to increase, having previously been de- 
creasing. In the former case u has a greater value for this 
particular value of x than it has for adjacent values of a?, and 
it is said to have a maximum value. In the latter case u has 
a less value for the particular value of x than it has for adja- 
cent values of x, and is said to have a minimum value. We 
say, then, that a function of a single variable is at a maximum 
value, or is a maximum, when it is greater than any of the 
values which immediately precede or follow it; and is at a 
minimum value when it is less than any of the values which 
immediately precede orfolloiv it. 

Thus, if we suppose the function u to be represented by the 
v2L variable ordinate PTl' of a curve 

line, and if the ordinate moving 
from A along AX gradually in- 
creases until it comes into the posi- 
: tion PR, and after that gradually 
decreases, the ordinate is said to be a maximum, or at a maxi- 
mum value, when it comes into the position PR 




MAXIMA AND MINIMA OF FUNCTIONS. 



95 



But if the ordinate moving along AX gradually decreases 
until it conies into the position pr, and after that gradually 
increases, the ordinate is said to be a minimum, or at a mini- 
mum value, when it comes into the position pr. 



80. Several maxima possible. These terms maxima and 
minima are not used to denote the greatest and least possible 
values which a function can assume, for the same function 
may admit of several maxima or several minima values, and 
some particular minimum may be greater than some particu- 
lar maximum of the same function. 
Thus, in the annexed figure. PR and 
MX represent two maxima values 
of the variable ordinate ; while pr, 
mn represent two minima values, 
and the minimum jp?* is greater than -£~J* ^~ 
the. maximum MN. 





81. Besides the case represented by the figure in Art. 80, we 
may have the case represented by the an- 
nexed figure, where PR is a maximum 
ordinate, because it is greater than the 
ordinate immediately preceding or fol- 
lowing; and jw is a minimum ordinate, 
because it is less than the preceding or 
following ordinate. 

82. How to discover maxima and minima values of a, func- 
tion. If we examine the figure Art. 79, we shall see that 
when x is increasing and y is approaching a maximum value 
as PR, the tangent to the curve makes an acute angle with the 

dv 
axis of X; that is, while approaching P from the left, y- is pos- 
itive. At P the tangent becomes parallel with the axis of X ; 

du 
that is, j- becomes zero. Immediately after passing P, the 



96 DIFFEKENTIAL CALCULUS. 

tangent to the curve makes an obtuse angle with the axis of 

v , . dy . 

X. ; that is, y- is negative. 

So, also, in approaching a minimum value as pr from the 

, „ dy . . . dy . _ 

leit, -j- is negative ; at the point jp, -y- is zero ; and after 

passings, -r- is positive. 

If we examine the figure in Art. 81, we shall see that while 
x increases and y is approaching a maximum value as PR, the 

dy 

tangent makes an acute angle with the axis of X ; that is, ~y 

is positive. At P the tangent in general becomes perpendicu- 

dv 
lar to the axis of X ; that is, -j- becomes infinite ; and after 

passing P, -j- is negative. 

Also, in approaching a minimum value as pr, -r- is negative ; 

at the pointy, -r- becomes infinite; and after passings, -~- is 
positive. 

Hence we see that for increasing values of the variable, 
when the function is a maximum the first differential coeffi- 
cient changes its sign from + to — ; and tohen the function 
is a minimum the first differential coefficient changes its sign 
from — to +. 

But a function which is continuous can change its sign only 
by becoming zero or infinity ; hence, for a maximum or min- 
imum value of a function, the first differential coefficient must 
he either zero or infinity. 

If, therefore, we find the first differential coefficient of the 
function, and put it equal to zero and equal to infinity, the 
roots of these equations will give all the values of x which 
can possibly render the function a maximum or a minimum. 

All the values of the variable thus found will not neces- 
sarily render the function a maximum or a minimum, because 
a variable quantity may pass through zero or infinity without 
changing its sign. Hence each value of the variable which 



MAXIMA AND MINIMA OF FUNCTIONS. 97 

renders the first differential coefficient zero or infinity must be 
tested separately. 

83. Mode of testing the roots thus obtained. Let a repre- 

dv 

sent one of the roots of the equation -j- = or oo. Substitute 

successively for x in the given function a + h and a— h, where 
h is supposed to be a quantity indefinitely small. If both these 
results are less than that obtained by substituting a, the root 
will render the function a maximum ; if both results are great- 
er, the root will make the function a minimum. 

If one result be greater and the other less than that obtained 
by substituting a, the root under consideration will give neither 
a maximum nor a minimum. 

Or we may substitute a + h and a — h successively in the first 
differential coefficient. If the first result be + and the second 
— , the root will render the function a maximum; if the first 
be — and the second + , the root will render the function a 
minimum. If the two results have the same sign, this root 
will give neither a maximum nor a minimum. 

The maximum or minimum value may be found by substi- 
tuting the corresponding root in the given function. 

Ex. 1. Find the value of x which will render u a maximum 
in the equation u = 10x—x 2 . 

Differentiating, we obtain 

^=10-2*. 

ax 

Putting this differential coefficient equal to zero, Ave have 

10-2^=0, 
whence x=5. 

Let us now substitute 5 — 1, 5, and 5 + 1 successively for x 
in the given function. 

Substituting 4 for #,we have ^ = 40—16 = 24; 
" 5 " " w =50—25 = 25; 

" 6 " " ^' = 60-36 = 24. 

Thus the function u appears to be a maximum when x = 5. 

E 



98 DIFFERENTIAL CALCULUS. 

This conclusion is not, however, a necessary inference from 
the above substitution, because the maximum value of the 
function might correspond to a value of x intermediate be- 
tween 4 and 5 or 5 and 6. In order to remove any uncer- 
tainty of this kind, we may proceed as follows : For x substi- 
tute successively 5— A and 5 + A, and in each case we obtain 
25 — A 2 . Now, however small A may be, so long as it has a 
finite value, 25— A 2 will be less than 25. Hence the value 
x=5 renders the function a maximum. 

Or we may substitute 5— A and 5 + A successively in the 
first differential coefficient. In the first case we obtain +2A, 
and in the second case — 2A. 

Ex. 2. Find the value of x which will render u a minimum 
in the equation u — x 2 — 16^+70. Ans. x=8. 

Ex. 3. Find the value of x which will render u a minimum 
in the equation u — 4 + (x — 3) 2 . A?is. x — 3. 

84. The first term of a series may he greater than the sum 
of all the others, A more convenient method of determining 
the values of x which will render any proposed function a maxi- 
mum or a minimum is founded upon the following principle: 

If the variable x receive a small increment A, and a func- 
tion ofx + hbe expanded into a series according to the ascend- 
ing powers of h, the quantity A may be assumed so small that 
any proposed term in the series shall be greater than the sum 
of all those that follow it. 

Let u / =u+ph + qh 2 + rh 3 + etc., (1) 

where p, q, r, etc., are functions of x independent of A. 

Equation (1) may be written 

u' = u + h(p + qh + rh 2 + etc.); 
or, putting T& = q + rh+ etc., we have 

Now, if A be indefinitely diminished, EA will become less than 
p, which is independent of h; that is,j9 is greater than RA, 
and consequently ph is greater than RA 2 , which is the sum of 
the succeeding terms of the series. 



MAXIMA AND MINIMA OF FUNCTIONS. 99 

In the same manner it may be proved that h may be taken 
so small that qh 2 shall be greater than the sum of the succeed- 
ing terms. This conclusion does not apply to the failing cases 
of expansion by Taylor's theorem, some of the coefficients in 
those cases becoming infinite. See Art. 65. 

85. Sign of the second differential coefficient examined. 
When the given function can be developed into a series by 
Taylor's theorem, the following method of finding maxima 
and minima will be most convenient : 

Suppose we have u=f{x), 

and let the variable x be first increased by h and then dimin- 
ished by A, and let 

a'=f{x + h), and u / =f{x—h)\ 
then, by Taylor's theorem, we shall have 

, du d 2 u h 2 d 3 u h 3 

u - u =Tx h +-M + ^ 1^3+ etc. ; (1) 

du 7 d 2 u h 2 d 3 u h? 

U - U = ~d^ h+ -d^ T^~W> 1^3+ etc. (2) 

Now, if u be a maximum, it must be greater than either v! or 

u n however small h may be assumed, and we have seen, Art. 84, 

du 
that h can be assumed so small that -y-h shall be greater than 

du 
the sum of all the terms which follow it ; hence, if -^ be 

either positive or negative, one of the quantities u', u / will be 
greater and the other less than ?/, which can not be the case if 
u be either a maximum or a minimum ; and hence, when u is 
either a maximum or a minimum, we must have 

du 
dx~ 
If the first differential coefficient vanishes, we shall have 
. d 2 u h 2 d 3 u h? < 

w - w= ^o+"^rx3+ etc - ( 3 ) 

d 2 n h 2 d 3 u h? ... 

U ~ U =-d^ ns"^ OT3+ etc - & 



100 DIFFERENTIAL CALCULUS. 

, . , , * \ . , , d 2 u h 2 

and since h can be assumed of such a magnitude that -t-j -r—9 

shall be greater than the sum of all the terms which follow it, 
and since h 2 is positive, the sign of the sum of each series will 
be the same as the sign of the second differential coefficient ; 
that is, u will be a maximum if the value of x determined 

from the equation -r-=0 renders the second differential coeffi- 

CLX 

cient negative, and a minimum if the second differential co- 
efficient is positive. 

If the second differential coefficient reduces to zero, the signs 
of equations (3) and (4) will again be opposite, and there can 
be neither a maximum nor a minimum unless the third differ- 
ential coefficient reduces to zero, in which case the sign of the 
sum of each series will depend upon that of the fourth differ- 
ential coefficient, and there will be a maximum when this is 
negative, and a minimum when it is positive, and so on. 

If the first differential coefficient becomes infinite instead 
of zero, this is one of the cases in which Taylor's theorem fails 
(Art. 65), and the demonstration which is founded upon it 
fails also. 

86. General rule for maxima and minima. Hence, to de- 
termine the value of x which will render a given function a 
maximum or a minimum, find the first differential coefficient 
of the function, place it equal to §) and find the roots of the 
equation thus formed. Substitute each of these roots in the 
second differential coefficient. Each one which gives a nega- 
tive result will, when substituted in the function) make it a 
maximum ; and each which gives a positive result will make 
it a minimum. 

If either root reduces the second differential coefficient to 
zerO) substitute in the third) fourth) etc.) until one is found 
which does not reduce to zero. If the differential coefficient 
which does not reduce to zero be of an odd order , this root 
will not render the function either a maximum or a mini- 



MAXIMA AND MINIMA OF FUNCTIONS. 101 

mum. But if it he of an even order, and negative, the root 
corresponds to a maximum ; if positive, to a minimum. 

Ex. 1. Find the values of x which will render u a maximum 
or a minimum in the equation 

u = x 3 -3x 2 -24x + 85. 

Differentiating, we obtain 

ax 
Placing this equal to zero, we have 

3x 2 - 6x— 24=0, 
or x 2 — 2x— 8=0, 

the roots of which are +4 and —2. 
The second differential coefficient is 

dx~ 2 = 6x - G - 
Substituting 4 for x in the second differential coefficient, 
the result is +18, which, being positive, indicates a minimum ; 
substituting —2 for x y the result is —18, which, being negative, 
indicates a maximum. 

Hence the proposed function has a maximum value when 
x— — 2, and a minimum value when x — ^. 

This result may be illustrated by assuming a series of values 
for x, and computing the corresponding values of u. Thus, 
If £=-4, u= 69; 
#=-3, ^ = 103; 
x—— 2, ^ = 113, maximum; 
x=— 1, ^=105 ; 
x— 0, u— 85; 
#=+1, u= 59; 
x=+2, u= 33; 
x— +3, t£=: 13; 
#=+4, w= 5, minimum; 
#= +5, u— 15; 
a?=r+6, w= 49. 
Thus it is seen that the value of the function increases 
while x increases from — 4 to — 2 ; it then decreases till x—^\ 



102 



DIFFERENTIAL CALCULUS. 




and after that it increases again uninterruptedly, and will con- 
tinue to do so, till x equals infinity. This peculiarity may be 
illustrated by a figure. 

If we assume the different values of x to represent the ab- 
scissas of a curve, and erect 
ordinates equal to the corre- 
sponding values of u, the curve 
line which passes through the 
extremities of all the ordinates 
will be of the form represented 
in the annexed figure, where 
it is evident that the ordinates 
show a maximum correspond- 
ing to the abscissa — 2, and a minimum corresponding to the 
abscissa 4. 

It will be noticed that in constructing the preceding figure, 
and also in several similar cases, the unit adopted for the ab- 
scissas is not the same as the unit adopted for the ordinates. 

Ex. 2. Find the values of x which will render u a maximum 
or a minimum in the equation 

u=x 3 -18x 2 + 96#-20. 
Ans. x=4: renders the function a maximum, and x =8 ren- 
ders it a minimum. 

Ex. 3. Find the values of x which will render u a maximum 
or a minimum in the equation 

v, = x 3 -18x 2 + 10$x. 
Ans. This function has a maximum value when x = 5, and a 
minimum value when x~7. 

Illustrate these results by a figure, as in the preceding ex- 
ample. 

Ex. 4. Find the values of x which will render u a maximum 
or a minimum in the equation 

u = x*—16x 3 + 88^ 2 -192^ + 150. 
The first differential coefficient gives us 

X 3 -12X 2 + 4:4:X-4:S = 0, 

the roots of which equation are 2,4, and 6. See Alg., Art. 464. 



MAXIMA AND MINIMA OF FUNCTIONS. 



103 



Substituting these values in the second differential coeffi- 
cient, we find that the given function lias a maximum value 
when #=4, and a minimum value when x = 2 or 6. 

If we assume a series of values for x, we shall obtain the 
corresponding values of u as follows : 
If a?=l, u= 31 ; 

6, minimum ; 



15; 

22, maximum ; 
15; 
6, minimum ; 



x = 2, u = 

x=3 y u = 

# = 4, u = 

x=5, u = 

x = 6, u = 

x=7, u= 31; 

x=8, u=150. 

The curve representing these 

values has the form represented 

in the annexed figure, where two 

minima are seen corresponding 

to the abscissas 2 and 6, and a 

maximum corresponding to the 

abscissa 4. 

Ex. 5. Find the values of x 

which will render u a maximum 

or a minimum in the equation 

700 
^r=^ 5 -25^ 4 + -g-^ 3 -1000^ 2 + 1920^-1100. 

Ans. This function has two maximum values corresponding 
to x = 2 and x = 6, and two minimum values corresponding to 
x=4t and x=8. 




87. Abbreviations of the process of finding maxima and 
minima. The process of finding maxima and minima may 
often be abridged by observing the following principles : 

(1) If the proposed function appears with a constant posi- 
tive factor, such factor may be omitted when seeking for those 
values of the variable which correspond to a maximum or a 
minimum. 



104 DIFFERENTIAL CALCULUS. 

Thus, let u — Ay, 

where y is a function of x, and A is any positive constant. 
By differentiating we have 

du Ady ^ d 2 u Ad 2 y 

dx"" dx } dx 2 ~~ dx 2 9 
from which it appears that those values of x which make 

du d 11 

-—- — will also make -7- = 0, and conversely. Also, any of 

dx dx ' J ' ■' 

those values which make -j-f negative will also make -j-r- 

dx 2 & dx 2 

ft it ft a 

negative; and any which make -t-| positive will make -ft 

positive. Hence every value of x which will make y a maxi- 
mum or a minimum will make Ay a maximum or a minimum. 
(2) Whatever value of x renders y a maximum or a mini- 
mum must render y n a maximum or a minimum, n being a 
positive integer. 

Let u=y n , 

y being any function of x. Then 

du ny n ~^dy 
dx~ dx y 

d u d 2 u dip" 

_ n y n -^ — ^ 4-n(n Y)v n ~ 2 

dx 2 ' & dx 2 v ;y dx 2 ' 

dy 
Now every value of x which will make ^- = will also make 

fin ffu 

-J-—0. Also (since -y— 2 = 0), if the same value makes ny n ~ l 

d 2 u d 2 y 

positive, it will render the sign of -i— the same as that of -r-f / 

that is, if it makes y a maximum or a minimum, it will make 
u a maximum or a minimum. If it makes ny n ~ l negative, it 

... . d 2 u . d 2 y _ . .. . 

will give to -ttt a S1 g n contrary to that or -f^/ that is, if it 

makes y a maximum, it will make u a minimum, and con- 
versely. 

The converse of this proposition does not hold true, that all 



MAXIMA AND MINIMA OF FUNCTIONS. 105 

the maxima and minima values of y n belong also to y. For 
the equation du n _ x dy 

d^ =ny dx=_ d 

may be satisfied by making either y n * = 0, or y- = 0. Those 

values of x which satisfy the first but not the second of these 
equations may make y n a maximum or a minimum, but will 
make y neither a maximum nor a minimum. 

Hence, since a radical expression is involved to the corre- 
sponding power by omitting the radical sign, in seeking for a 
maximum or a minimum of a function which is under a 
radical sign, the radical sign may be omitted. Care must be 
taken, however, not to use any of those values which belong 
only to the power. 

(3) In a similar manner it may be shown that in general 
any value of the variable which will render u a maximum or 
minimum will also render log. u and a u a maximum or mini- 
mum ; and also any value which will make u a maximum will 

make — , as also — Au, a minimum, and conversely. 

Ex. 1. Find the values of x which will render u a maximum 
or a minimum in the equation 

u = 3x 3 ~ x + c. 

Ans. X—- renders the function a minimum; 

1 

and x— — o renders it a maximum. 

Ex. 2. Find the values of x which will render u a maximum 
or a minimum in the equation 

u — x 3 — 9# 2 + 15#— 3. 
Ans. x — \ gives a maximum, and x=5 a minimum. 

Ex. 3. Find the values of x which will render u a maximum 
or a minimum in the equation 

U — zrrr- . 

# — 10 

Ans. x=4: gives a maximum, and x = 16 a minimum. 
E2 



106 



DIFFERENTIAL CALCULUS. 



Ex. 4. Find the values of x which will render u a maximum 
or a minimum in the equation 

1—x+x 2 



u- 



'1+x—x 2 ' 



Ans. x=n gives a minimum. 

Ex. 5. Find the values of x which will render u a maximum 
or a minimum in the equation 

u=x 3 -3x 2 + 6x+7. 

Ans. This function has neither a maximum nor a minimum. 

Ex. 6. Find the values of x which will render u a maximum 
or a minimum in the equation 

u-- 



f,=V< 



x° 



■3x 2 . 



Ans. x=0 gives a maximum, and x=2 gives a minimum. 



GEOMETRICAL PROBLEMS IN MAXIMA AND MINIMA. 

88. General directions. In the solution of geometrical prob- 
lems in maxima and minima, we must first obtain an algebraic 
expression for the quantity whose maximum or minimum state 
is required. This expression must be derived from the known 
properties of the figure in question, and will, of course, involve 
the variable or variables on which it depends. If the expres- 
sion contains two dependent variables, we must find, from the 
known properties of the figure, or from the conditions of the 
problem, the relation between these variables, and substitute 
for one of them its value in terms of the other. We shall 
thus obtain a function of a single variable, and we may then 
proceed as with the algebraic problems of Art. 83, etc. 

Prob. 1. It is required to find the maximum rectangle which 
C can be inscribed in a given triangle. 

Let b represent the base of the triangle 

ABO, h its altitude, and x the altitude of 

the inscribed rectangle. Then, by similar 

triangles, we have 

CD:CG::AB:EF, 
or h:h-x::i:EF. 



E 



D 



GEOMETRICAL PROBLEMS IN MAXIMA AND MINIMA. 107 

Hence EF=|(A^a>). 

The area of the rectangle is equal to EF x GD, 
or j(hx—x 2 ), 

which is to be a maximum. 

But, since t is a constant, the quantity hx—x 2 must also be a 
maximum (Art. 87). 

Hence -j-=/i — 2x=0, 

ax 7 ' 

or x= q. 

Also, since ~j~^— —2, we know that the value x—-^ gives a 

maximum value of the rectangle ; that is, the altitude of the 
maximum rectangle is equal to half the altitude of the tri- 
angle. 

Prob. 2. It is required to find the sides of the maximum 
rectangle which can be inscribed in a circle whose radius is r. 

Ans. Each is equal to r^/2. 

Prob. 3. It is required to find the sides of the maximum 
rectangle which can be inscribed in a semicircle whose radius 

Ans. ?* -y/2, and — ^. 

Prob. 4. It is required to find the sides of the maximum 
rectangle which can be inscribed in an ellipse whose axes are 
2a and 2b. 

The equation of an ellipse is 

b 

y=-Va 2 -x 2 . 

The area of the rectangle 

^bx ^b 

Hence a 2 x 2 —x* must be a maximum. 

Ans. a^/2, and b^/2. 



108 



DIFFERENTIAL CALCULUS. 



Prob. 5. It is required to find the sides of the maximum 
rectangle which can be inscribed in a semi-ellipse ; 1st, when 
the bisecting line is the major axis ; 2d, when it is the minor 
axis. j a 

Ans. a^/2, and -j^\ —7^, and b\/2. 

Prob. 6. What is the altitude of the maximum rectangle 
which can be inscribed in a segment BAC of a parabola ? 

Put AD = A, and AE=x; then, by 
the equation of the parabola y 2 — ^ax^ 
we have GE 2 :=4a#. 
Hence 

GE = 2 Vox, and GH = 4 Vox. 
Therefore, the area of GHKI is 

which is a maximum. 
xf is a maximum. 
h 




or 



x{h- 



Hence 



x- 3 . 



Or, the altitude of the maximum rectangle is two thirds of the 
axis of the parabola. 

Prob. 7. Required the least square which can be inscribed 
in a given square. 

Ans. Each angle of the required square is on the 
middle of a side of the given square. 

Prob. 8. Within a given parabola BAC to inscribe the great- 
est parabola EDF, having its vertex D in 
the middle of the base BC. 

Let AT)=h, and AG=#; then 

EG 2 =4ax, and EF=4\/aa. 
Therefore the area of the parabola 

edf=|efxgd, 

2 — 
= o X 4:Vax(h — x), a maximum ; 




GEOMETRICAL PROBLEMS IN MAXIMA AND MINIMA. 



109 



or 
Hence 



x(h—x) 2 is a maximum. 



DG = 



TjA, Arts. 



Prob. 9. What is the length of the axis of the maximum 
parabola which can be cut from a given right cone ? 
Put BC = J, AB = A, and CE=x; then BE = b-x. 
Also (Geom., Bk. IV., Pr. 23), 

FE=Vbx-x 2 , and TG = 2Vbx-x 2 . 
By similar triangles we have 

£:A::z:DE = y. 

Hence the area of the parabola is 
2 hx 



Hence 
and 

Also, 



o . -T- . 2 V to — # 2 , a maximum. 
xVbx—x 2 is a maximum, 



be=^=?; 




6~4 

that is, the axis of the maximum parabola is three fourths of 
the slant height of the cone. 

Prob. 10. It is required to circumscribe about a given parab- 
ola an isosceles triangle whose area shall be a minimum. 

Put a>=MP=AM; y=NP; A=MD; 

then y 2 = 4a#, or y = 2 -/«#. 
Also AP:NP::AD:BD, 

(x+h)2Vax 



or 2x:y: \x + h\ BD = 

Area of , 7NO /— 
{x+Kfvax 



2x 



ABC: 



cc 



= a minimum. 




-r # 2 + 2#A + A 2 411 

Hence ; =^ + 2Ao? 2 -fA 2 ^ 2 =a minimum. 

TT VX h A AT* 4A 

Hence a? = ~ A, and Al) = -=r ; 

that is, the altitude of the triangle is four thirds of the axis of 
the parabola. 



110 



DIFFERENTIAL CALCULUS. 




Prob. 11. What is the altitude of a cylin- 
der inscribed in a given right cone, when the 
volume of the cylinder is a maximum ? 

Let h represent the height of the cone, 
h the radius of its base, and x the altitude of 
the inscribed cylinder. Then, by similar tri- 
angles, we have AD : BD : : AE : EF, 

or h\b\\h— #:EF = t(A— x). 

The area of the circle whose radius is EF is 

^{h-xf. 

Multiplying this surface by DE, we obtain the volume of the 
cylinder T2 

-p— (A—x) 2 , a maximum. 

Hence x=-k] that is, the height of the greatest cylinder is 

one third of the altitude of the cone. 

Prob. 12. It is required to determine the dimensions of a 
cylinder which shall have a given volume with the least pos- 
sible surface. 

Let c denote the volume of the cylinder, and x the radius 
of the base ; the area of the base will be represented by nx 2 . 



irx< 



Hence the height of the cylinder equals 

The convex surface of the cylinder is 

c ^ 2c 
nx d x 

Hence the entire surface of the cylinder is 

2c 



— + 27rx 2 = a minimum. 
x 



Hence 



and the height of the cylinder 



GEOMETRICAL PROBLEMS IN MAXIMA AND MINIMA. Ill 

that is, the height of the cylinder is equal to the diameter of 
the base. 

Prob. 13. It is required to determine the dimensions of a 
cylindrical vessel open at top which has the least surface with 
a given capacity. 

Ans. The altitude of the cylinder is equal to the radius of 
the base. 

Prob. 14. Required the altitude of the greatest cylinder 
which can be cut out of a sphere whose diameter is D. 

Ans. Dy 3. 

Prob. 15. Required the maximum cone which can be in- 
scribed in a given sphere. B 

Let AC=r2^ and AD=#. 
Then BD 2 =AD xDC = 2ra-cc 2 . 

7TX 

The solidity of the cone = -5- (2rx — x 2 ), a| 

a maximum. 

Hence m=—; 

that is, the altitude of the cone whose volume is a maximum 
is four thirds of the radius of the sphere. 

Prob. 16. Required the altitude of a cone inscribed in a 
given sphere which shall render the convex surface of the 
cone a maximum. 

Ans. Four thirds of the radius of the sphere. 

Prob. 17. It is required to determine the dimensions of a 
cone which shall have the greatest volume with a given 
amount of surface. 

Let # = the radius of the base, and ?/ = the slant height. 
The total surface of the cone = 7r# 2 + 7r#y=<?. 

Hence y— — —x. 

* 7TX 




Solidity of the cone = o7r^ 2 \/ 



2c 



n 2 x 2 it 



or — —2a? 4 is a maximum. 

7T 



112 DIFFEKENTIAL CALCULUS, 

Hence =5\/? 

3 fc 



'2Vtt' 



and y-. 

that is, the slant height is three times the radius of the base. 

Prob. 18. It is required to determine the dimensions of a 
cone whose convex surface is given (=c), and whose volume 
is a maximum. 






Ans. The radius of the base = \/- 



71-^/3 

Prob. 19. It is required to determine the altitude of the 
least cone circumscribed about a given sphere. 

Ans. Twice the diameter of the sphere. 
Prob. 20. What is the altitude of the maximum cylinder 
which can be inscribed in a given paraboloid ? 

Ans. Ilalf the axis of the paraboloid. 



TANGENTS AND NORMALS TO PLANE CURVES. 113 



CHAPTER V. 

TANGENTS AND NORMALS TO PLANE CURVES. ASYMPTOTES. 

89. Equation of a tangent line. If x', y' represent the co- 
ordinates of a given point on a given curve, the equation of a 
straight line passing through this point will be (Anal. Geom., 
Art. 38) y-y'=im(x-x), (1) 

where m represents the tangent of the angle which the line 
makes with the axis of abscissas. But we have seen, Art. 77, 
that the tangent of the angle which a tangent line at any point 
of a curve makes with the axis of abscissas is equal to the first 
differential coefficient of the ordinate of the curve ; hence, in 

dy f 
place of m in equation (1) we may substitute y- ,, and we ob- 

tain y-/=^(*--aO, ( 2 ) 

which is the equation of a tangent to a curve at a point whose 
co-ordinates are x', y\ 

When it is required to find the equation of the tangent line 
to any curve, we must differentiate the equation of the curve, 

and find the value of -j- n which is to be substituted in equa- 
tion (2). dX 

Ex. 1. Let it be required to find the equation of the tangent 
line to a circle. 

The equation of the circle is 



and by differentiation we 


y* = <) 
) find 


a -x 2 , 










Substituting this value 


dy' x' 
dx'~ y'' 
in equation 


(2), we 


have 


for the 


equa- 


tion of a tangent line 




x' 

-—(X- 










y- 


y'=- 


-x'), 









114 DIFFERENTIAL CALCULUS. 

whence xx' + yy f = x' 2 + y' 2 = r 2 , 

which corresponds with Anal. Geom., Art. 73. 

Ex. 2. Find the equation of the tangent to an ellipse. 
The equation of an ellipse is 

a 2 y 2 =a 2 b 2 — b 2 x 2 , 
dy' b 2 x f 

whence m=~w 

Hence the equation of the tangent line is 

or a?yy r + b 2 xx' — a 2 y /2 + b 2 x' 2 — a 2 b 2 , 

which corresponds with Anal. Geom., Art. 134. 

Ex. 3. Find the equation of a tangent line to a parabola. 

Ex. 4. Find the equation of a tangent to the curve repre- 
sented by 3y 2 +2x 2 = 6 

at the point x=l, and construct the figure. Is there more 
than one tangent ? 

90. Equation of a nor?nal line. Since the normal is per- 
pendicular to the tangent, the equation of a normal at the 
point x\ y r is -i 

y-y>= --<*-<*), (i) 

dx' 

or y-y'=-fyfa-x')- ( 2 ) 

Ex. 1. What is the equation of a normal to the curve repre- 
sented by 3x 2 + 4y 2 =zl2 
at the point x = 11 

Ans. At 8=1, y==fcf< Hence there are two points indi- 
cated. The equation of the normal at the former 
is y— 2x + \ — 0\ and at the latter y + 2#— ^=0. 
Construct the figure and the two normals. 

91. The subtangent of any curve is the portion of the axis 
of abscissas intercepted between the tangent and the ordinate 
drawn from the point of tangency. 



TANGENTS AND NORMALS TO PLANE CURVES. 115 

If in equation (2), Art. 89, we make y=zO,the corresponding 

dx' 
value of x' —x, which is y'^T 7 * 

will denote the length of the subtangent, where x\ y' is the 
point of tangency. 

The same result may be deduced from v^ 

the annexed figure. Let PT be the tangent 
cutting the axis of abscissas in T. Then in sy 

the right-angled triangle PTR we have y* / 

tang.T:PR::l:TR; T ~ 

or, the length of the subtangent to any point of a curve re- 
ferred to rectangular axes is equal to the ordinate multiplied 
by the differential coefficient of the abscissa. 

92. Length of the tangent. In the right-angled triangle 
PTR we have PT 2 = PR 2 + RT 2 ; 

dx 
that is, PT 2 =^+^. 



/ iP'dx^ I dx^ 

Hence the tangent PT = \/y*+^=y\/ 1 + df' 

The same result may be expressed thus : The length of the 
tangent to any point of a curve referred to rectangular axes 
is equal to the square root of the sum of the squares of the 
ordinate and subtangent. 

93. The subnormal of any curve is the portion of the axis 
of abscissas intercepted between the normal and the ordinate 
drawn from the same point of the curve. 

If in equation (2), Art. 90, we make y = 0, the corresponding 
value of x—x', which is t dy' 

will denote the length of the subnormal, where x', y' is the 
point of the curve from which the normal is drawn. 



116 DIFFERENTIAL CALCULUS. 

^/i The same result may be deduced from 

s/\ the annexed figure. In the right-angled 

// \ \ triangle PRN we have the proportion 

// \ \ 1 : PR : : tang. RPN : EN. 

T R sr But the angle RPN is equal to PTR; 

hence 1 : PR : : tang. PTR : RN ; 

that is, l:y::|:K]Sr^|; 

or, the length of the subnormal to any point of a curve re- 
ferred to rectangular axes is equal to the ordinate multiplied 
by the differential coefficient of the ordinate. 

94. Length of the normal. In the right-angled triangle 
PRN we have PN 2 = PR 2 + RN 2 ; 

that is, VW=y*+^f; 

Hence the normal F^=\/y 2 +^^=y\/l+^ 2 . 

The same result may be expressed thus : The length of the 
normal to any point of a curve is equal to the square root of 
the sum of the squares of the ordinate and subnormal. 

95. Tangent, etc., of a particular curve. To apply these 
formulas to a particular curve, we must substitute in each of 

doc dii 
them the value of -y- or -7- obtained by differentiating the 
dy dx J -° 

equation of the curve. The results thus obtained will be gen- 
eral for all points of the curve. If the values are required for 
a particular point of the curve, we must substitute in these re- 
sults for x and y the co-ordinates of the given point. 

Let it be required to apply these formulas to lines of the 
second order whose general equation is 

y 2 — mx + nx 2 . 
Differentiating, we have 

dy m + 2?ix m + %\x 
dx *2y 2Vmx + nx 2 



TANGENTS AND NORMALS TO PLANE CURVES. 117 

Substituting this value in the preceding formulas, we find 

dx 2{mx-\-7ix 2 ) 

the subtan °;ent = y-j- = r~a 5 

& J ay m + znx ' 

/ y 2 dx 2 [~ /?nx + 9ix 2 \ 2 

the tangent=V2/ 2 + ^r=V mx + nx+ \^^) 5 

the subnormal = y-^- = ~ — ; 



the normal = y y 2 + —j - ^ - = V^a? + nx 2 + \(7ii + 2;*#) 2 . 

96. Tangent, etc., to a parabola. By attributing proper 
values to m and ?i, the above formulas will be applicable to each 
of the conic sections. For the parabola n = 0, and these ex- 
pressions become, 

the subtangent = 2#, which corresponds with Anal. Geom., 

Art. 93 ; 
the tangent = -y/mx + ±x* ; 

7?? 

the subnormal = -^ , which corresponds with Anal. Geom., 

Art, 98; 



the normal i , 



f~ m 2 
= y mx+-r 



97. Subtangent, etc., of an ellipse. In the case of the ellipse, 
these expressions assume a simpler form when the origin of co- 
ordinates is placed at the centre. The equation then becomes 

a 2 ]/ 2 + b 2 x 2 = a 2 b 2 ; (1) 

whence, by differentiating, we obtain 

dy b 2 x 
dx~~ a 2 y 

Hence, from Art, 91, we find „ , 

a ij 
the subtangent of the ellipse ——-—-. (2) 

But from equation (1) we have 



a 2 y 2 



118 DIFFERENTIAL CALCULUS. 

Hence, from equation (2), we find 



the subtangent of the ellipse : 



x ? 

which corresponds with Anal. Geom., Art. 135. 

Also, from Art. 93 we find ? t 2 

the subnormal of the ellipse ——r- — _ — 

r ax ar' 

which corresponds with Anal. Geom., Art. 139. 

98. Subtangent, etc., for a circle. In the case of the circle, 
a and b become equal, and we find 

the subtangent of the circle = — — : 
& x ' 



the tangent of the circle = \] if + ^ = y^-f- = ^; 

X XX 

the subnormal of the circle =—x: 



the normal of the circle = Vy 2 + x 2 =r 2 ; 
which results accord with well-known principles of Geometry. 

Ex. 1. The latus rectum of a parabola is 4 inches, and an 
abscissa 9 inches; required the length of the corresponding 
ordinate and subtangent. Ans. 

Ex. 2. The major axis of an ellipse is 30 inches, and the 
minor axis 16 inches; required the length of the subtangent 
corresponding to an abscissa of 10 inches, measured from the 
centre. Ans. 

Ex. 3. The major axis of an ellipse is 6 inches, and the 
minor axis 4 inches ; required the length of the subnormal 
corresponding to an abscissa of 2 inches, measured from the 
centre. Ans. 

Ex. 4. The diameter of a circle is 10 inches ; what is the 
length of the tangent and subtangent corresponding to an ab- 
scissa of 3 inches, measured from the centre ? 

Ans. 

Ex. 5. What is the value of the subtangent and subnormal of 
the curve represented by y 3 = axf 

Ans. Subtangent = 3x ; Subnormal = g-. 






ASYMPTOTES TO CURVES. 



119 



Ex. 6. Required the value of the sub tangent of the loga- 
rithmic curve. 

The equation of the logarithmic curve (Anal. Geom., Art. 
215) is &=log.y. 

Differentiating, Art. 36, we obtain 



or 



y 

dy- 




or, the subtangent of the logarithmic 
curve is constant, and equal to the 
modulus of the system of logarithms 
employed. 

In the Naperian system M equals 
unity, and hence the subtangent AR is equal to unity, or 
AB. It is obvious that if the point of tangency be taken 
either to the right or left of P, as shown in the figure, the tan- 
gent line will not pass through the origin. 

ASYMPTOTES TO CURVES. 

99. Branch of a carve. Let x be the abscissa of any curve, 
and y the corresponding ordinate, and let 

be the equation of the curve. If we assign to x arbitrary 
values, and find the corresponding values of y, we shall de- 
termine different points of the curve, and by this means the 
curve can be constructed. If for each positive value of x the 
equation gives two values for y, this indicates that the curve 
has two branches, and if the values of y are all real when the 
assumed values of x increase indefinitely, the branches of the 
curve extend indefinitely ; but if the values of y become 
imaginary, this indicates that the curve does not extend indef- 
initely on the positive side of the axis of ordinates. In the 
same manner, by assigning to x negative values, we may de- 
termine whether the curve has branches extending indefinitely 
on the negative side of the axis of ordinates. The parabola 
has two branches extending indefinitely on the positive side of 



120 DIFFERENTIAL CALCULUS. 

the axis of ordinates, but none on the negative side. The hy- 
perbola has two branches extending indefinitely on either side 
of the axis of ordinates. 

In Analytical Geometry, Part II., Section IX., are given the 
equations of numerous curves which have branches extending 
indefinitely in various directions. 

Ex. 1. Determine whether the curve y — x 3 —x has an infinite 
branch, and construct the figure. 

Ex.2. Determine w T hether the curve yx 3 = 81 has an infinite 
branch, and construct the figure. 

100. Asymptotes. Suppose a curve has one or more branch- 
es extending to an infinite distance from the origin, and that 
tangent lines are drawn at successive points of such a branch. 
Then, as we pass along the curve from point to point, we some- 
times find that the directio?is of these tangents approach some 
definite limit, and also that the intercepts cut from the axes of 
co-ordinates tend to a finite limit. In such a case, there is a 
straight line tow r ard which the successive tangents continually 
approach, and this straight line is called an asymptote to the 
curve. An asymptote to a curve is the limiting position of 
the tangent when the point of contact moves to an infinite 
distance from the origin; or we may define an asymptote to 
be a line which will not meet the curve at any finite distance 
from the origin, hut to which the curve may he made to ap- 
proach as near as we please by continually increasing or de- 
creasing one of the co-ordinates. 

101. Illustration. Suppose a series of ordinates, BM, CN, 
DO, etc., to be drawn at equal distances 
from each other, and let each succeed- 
ing ordinate be half of the preceding. 
The curve which passes through the ex- 
tremities of all these ordinates will 
continually approach the line AX, but 
will not m.eet it at any finite distance 




ASYMPTOTES TO CURVES. 



121 



from A. The line AX is called an asymptote to the curve 
MNP, and is regarded as a tangent to the curve at an infinite 



distance from the origin. 



Asymptotes may be either curvilinear or rectilinear, 
shall consider-only the latter. 



We 



102. To determine whether a plane curve has rectilinear 
asymptotes. First determine whether the curve has one or 
more branches extending to an infinite distance from the 
origin. If for all positive values of x from -\-a to +00 one or 
more values of y are real, then for each set of real values of y 
the curve has an. infinite branch extending to the right ; if for 
all negative values of x from —a to —00 one or more values 
of y are real, then for each set of real values of y the curve 
has an infinite branch extending to the left. If the curve has 
no infinite branch, it can not have an asymptote, since an 
asymptote is a tangent at an infinite distance from the origin. 

If the curve has an infinite branch, we may proceed as fol- 
lows to determine whether there is an asymptote to that branch: 

Let MPQ be any plane curve, and 
let PT be a tangent to the curve at 
the point P whose co-ordinates are 
AR=^ and PR = ?/7 and let PT 
meet the two axes in B and T. The 
equation of PT, Art. 89, is 

y-y =fa( x - x )- c T 

If in this equation we make y=0, we obtain 

y'dx' 




If we make x- 



aj=AT=#' 

:0, we obtain 

y=AB=y' 



■ dy" 

x'dy' 

dx 



(i) 

(2) 



Xow, when the point of contact P recedes from the origin, 
the tangent PT approaches the asymptote KDC ; and when P 
reaches an infinite distance, the tangent coincides with the 
asymptote, and AT becomes AC, and AB becomes AD. 



122 DIFFERENTIAL CALCULUS. 

If, when one or both of the co-ordinates x' and y f are indef- 
initely increased, the intercepts AD and AC are both finite, 
there will be an asymptote, whose position is determined by 
the two points D and C. 

If one of these intercepts becomes infinite, and the other 
finite, there will be an asymptote parallel to that axis on which 
the intercept is infinite. 

If both intercepts become infinite, there will be no asymptote. 

If both intercepts become zero, the asymptote will pass 
through the origin of co-ordinates, and the tangent of the angle 
which it makes with the axis of X will be given by the expres- 

dy f 

sion -5-7, when one or both of the co-ordinates become infinite. 

Ex. 1. It is required to determine whether the hyperbola has 
asymptotes. 

The equation of the hyperbola when the origin of co-ordi 
nates is at the centre is (Anal. Geom., Art. 170) 

b 2 

When x is greater than a, y has always two real values, how- 
ever great x may be, and hence there are two infinite branches 
to the right. 

Differentiating, we find 

dx a 2 y 2 x 2 — a 2 
dy~ b 2 x " x 

Therefore, A T=x'-^-~. 

CL1J X 

Also, n^ -£ 

d 2 

When x is supposed to be infinite, the intercept AT = — be- 

x 

comes zero. Also, when x is infinite, y must be infinite, and 

b 2 

the intercept AB= — — becomes zero. 

Hence the hyperbola has two asymptotes, which pass through 
the origin, or the centre. 






ASYMPTOTES TO CURVES. 123 

To determine the direction of the asymptotes, we have 
dy b 2 x bx 

dx~~~a 2 y aVx 2 -a 2 ' 

When x— + 00, this expression = db— . Hence we see that 

the asymptotes are the produced diagonals of the rectangle 
described upon the axes. 

In the direction of x negative there are also two infinite 
branches, which will be found to have the same lines through 
the origin as asymptotes. In fact, it will be found that every 
asymptote to an algebraic curve is an asymptote to two branch- 
es of the curve. 

Ex. 2. It is required to determine whether the parabola has 
asymptotes. 

The equation of the parabola is 

y^ — ^ax. 

We see that there are two infinite branches to the right. 

Differentiating, we find 

dx y 2 
^dy~2a~ 

Hence AT=zx— ^-j- —— x. 

dy 

When x is supposed to be infinite, AT = — o> ; and when y 
is infinite, AB= + 00. Hence the parabola has no asymptotes. 

The ellipse and circle can have no asymptote, since neither 
has an infinite branch. Hence the hyperbola is the only line 
of the second order which has asymptotes. 

Ex. 3. It is required to determine whether the logarithmic 
curve has asymptotes. 

The equation of the logarithmic curve is 

x=\og.y, 

oi- y—^- 



124 



DIFFEKENTIAL CALCULUS. 




i-C 



If x be taken infinite and 
negative, then 

that is, the axis of abscissas is 
an asymptote to the curve. 

This curve has also an in- 
finite branch extending to the 
right, to which there is no 
asymptote. 
Ex. 4. It is required to determine whether the curve repre- 
sented by y 3 = 10 — x 3 
has asymptotes. 

This curve has an infinite branch ex- 
tending to the right, and also one ex- 
tending to the left. 

Differentiating, we find 

AT =— , andABrr— , 
x y 

each of which reduces to zero when x and y are infinite. Hence 
the curve has an asymptote passing through the origin. 

The direction of the asymptote is given by the expression 




dy 
dx 



X 2 (lO-y 3 ) 1 



, which equals — - 1 when y is infinite. 



r y 

Hence the asymptote makes an angle of 135° with the axis of X. 
Ex. 5. It is required to determine whether the curve repre- 
sented by if=x 3 + x 2 
has asymptotes. 

Differentiating, we find 




AT: 



~3x + # 



--£• when x is infinite. 



3(x 3 + x 2 )& 
which =■£ when x is infinite. 



TANGENTS TO CURVES REFERRED TO POLAR CO-ORDINATES. 125 

Hence there is an asymptote which cuts the axis of Y at -J 
above the origin, and the axis of X at -J- on the left of the 
origin. The asymptote therefore makes an angle of 45° with 
the axis of X. 

TANGENTS TO CURVES REFERRED TO POLAR CO-ORDINATES. 

103. To find the angle included between a tangent to a 
polar curve and the radius vector draion to the point of con- 
tact. Let P be the pole of the curve, and 
let the straight line MS cut the curve in 
two points, M and N. Represent the ra- 
dius vector PM by r, PN by r\ the angle 
XPM by 0, and the angle MPN by h. 
Also draw NR perpendicular to PM pro- 
duced. Then we shall have i/ /s 

mf , OT ■ NR NR PNsin.A 

taiig.PM8 = tang^ME== ra =p SZ p H == pN ^ A-pM = 

r r sin. h 
r r cos. h—r ^ ' 




But cos. A = l--2(sin.^A) 2 , and sin.A = 2 sin.-JA cos.JA 
nee 
tang. PAIS 



Hence 

r sin. A 



*&• 



r / -r^2r\sm.ihf-r / ~r , ' (2) 

— — 7 —r tang, i-h. 
sin. A & z 

If now the point N be supposed to move toward M until 
it coincides with it, the secant line MS will come into the posi- 
tion of the tangent MT at M, and tang. PMS will become 
tang. PMT, which is therefore equal to the limiting value of 
the second member of equation (2), when h is indefinitely 

rp ry% fir 

diminished. But the limit of - — 7 is -375 »' and the limit of 

7 sm. h dv' 

h . TT 

tang. 7> is zero. Hence 

r rdO 
or, the tangent of the angle which a tangent line to the curve 




126 DIFFERENTIAL CALCULUS. 

forms with the radius vector is equal to the radius vector 
multiplied by the differential coefficient of the variable angle. 

104. A polar subtangent is a line drawn from the pole per- 
pendicular to a radius vector, and limited by the tangent 
drawn through the point where the radius vector meets the 
curve. 

Thus, if P be the pole of the curve, 
and MT be a tangent to the curve at the 
point M, then PT drawn perpendicular 
to PM, and meeting the tangent in T, is 
~t called the polar subtangent. 
A polar subnormal is a line drawn from the pole per- 
pendicular to a radius vector, and limited by the normal 
drawn through the point where the radius vector meets the 
curve. 

Thus, if ML be drawn perpendicular to the tangent MT, 
and meet the line LPT drawn perpendicular to MP in L, then 
PL is called the polar subnormal. 

105. To find the value of the subtangent : , etc., to a 

curve referred to polar co-ordinates. The subtangent PT = 

rdf) r^df) 
PM tang.PMT = r-^-=:— *— / or, the length of the subtangent 

to a curve referred to polar co-ordinates is equal to the square 
of the radius vector multiplied by the differential coefficient 
of the variable angle. 

t / rV/ft 2 / r 2 df) 2 

Tangent MT= VFW+FP=\/** +-££- = 'V^^pr * 

or, the length of the tangent in polar co-ordinates is the square 

root of the sum of the squares of the subtangent and radius 

vector. 

The subnormal PL=PM tang.PML=:PM cotang.PMT = 

rcJ r (I r 

—jt:=-jj:; or, the subnormal to a polar curve is equal to the 

differential coefficient of the radius vector. 



TANGENTS TO CURVES REFERRED TO POLAR CO-ORDINATES. 127 

Ex. 1. Find the value of the subtangent and subnormal to 
the spiral of Archimedes. -, n 

The equation of the spiral of Archimedes is r=K~ ; whence 

^_2tt n 

dr~~ b ' 

r 2 dd 2irr 2 2tt bW bO 2 

Hence the subtangent =— j— = — r~ =~r* ~x~ T z—Tr~' 
& dr b b 4tt 2 2ir 

If the moving point is supposed to start from the pole, then 
at the end of one revolution of the radius vector = 27r, and 
the subtangent = 2#7r, which equals the circumference of the 
circle described with the radius vector of the point of contact. 

If the tangent be drawn at the extremity of the arc gen- 
erated in m revolutions, we shall have = 2ni7r, and the sub- 

bd 2 Abm 2 * 2 rt7 ^ 7 , 

tangent =-r-= — — 2bmH — m.2bmir\ or, the subtangent 

after m revolutions is equal to m times the circumference of 

the circle described with the radius vector of the point of 

contact. n 7 

The subnormal — -jt,— 77-. 
dO 2?r 

Ex. 2. Find the value of the subtangent of the hyperbolic 

spiral. 

The equation of this spiral is ^ = -/ 

dO 2 
whence -7- = — —. 

dr a 

„ , , r 2 dO r 2 2 

Hence the subtangent — —1 — = — = —a; 

to dr a ' 

that is, in the hyperbolic spiral the subtangent is constant. 

Ex. 3. Find the value of the subnormal of the logarithmic 

spiral. 

The equation of this spiral is 6=log.r; whence 

dQ_M 

dr~ r' 

dr r 
Hence the subnormals ~jh—Tj[i where M represents the mod- 
ulus of the system of logarithms employed. In the Naperian 



128 DIFFERENTIAL CALCULUS. 

system of logarithms, M equals unity ; hence in the JVaperian 
logarithmic spiral the subnormal is always equal to the ra- 
dius vector. 

rdO 
We also have tang. FMT=-^=M; that is, the tangent of 

the angle which the tangent line makes with the radius vector 
is constant, and is equal to the modulus of the system of loga- 
rithms employed. In the Naperian logarithmic spiral the tan- 
gent line is inclined 45° to the radius vector. 



CONVEXITY AND CONCAVITY. 



129 



CHAPTER VI. 

CONVEXITY AND CONCAVITY. SINGULAR POINTS OF CURVES. 
TRACING OF CURVES. 

106. A curve is said to be convex at one of its points with 
respect to a given line, when in the immediate vicinity of that 
point the tangent lies between the curve and the given line. 
The curve is said to be concave when it lies between its tan- 
gent and the given line. 



107. To determine when a curve is convex toward the axis 
of abscissas. 

Let PPT" be a curve convex to- 
ward the axis of abscissas and above 
that axis, and let x and y be the co- 
ordinates of the point P. Let x be 
increased by any arbitrary increment 
RE/, which we will represent by A, 
and take R'R" also equa] to A. Draw 
the ordinates RT',K"P"; draw the 
secant line PP', and produce it to B ; 
join PT ;/ , and draw PD, P'D' par- 
allel to AR. We shall then have 

RP=/( aJ )=yy 

dy d 2 y h? 









F 


7 


5^ 


c 


B 
D' 


p^^ 


D 
R' 




R 


R" 






d 




p 


\ 


d' 


P 


\ 


%: 


b 








24 

prr 


\ 



KT>=f(x+h)=y + /h + -^ 2 ^+ etc 



dx ^ dx 2 2 
dy d 2 y 4A 2 



Also, B,'T"=f(x+2h)=y+-£2h+ 



dx ^ dx 2 2 



+ etc. 



Hence 



DP-E'F-EP^A+gf+ete. 



Also, DT"=R"P' 



■ B ' F =dx h+ d^^ +etc - 
F2 



(1) 
(2) 
(3) 
(4) 



130 



DIFFERENTIAL CALCULUS. 



Subtracting equation (3) from equation (4), remembering 
that D'B is equal to DP', we have 



BF 



72 

:DT".-DP=y^ 2 + etc. 



(5) 



Now, Art. 84, the increment h can be taken so small that the 
sign of the second member of equation (5) will be the same as 
that of its first term ; and since, by the definition of a curve 
convex toward the axis of a?, the first member of the equation 
is positive, the second must also be positive ; that is, the second 
differential coefficient is positive, and the ordinate, being situ- 
ated above the axis of abscissas, is also positive. 

If the curve is below the axis of abscissas, we shall have 

ty'=dy'-dp'=^h*+ etc.; 

and since the first member of this equation is negative, the 

second will also be negative ; that is, the second differential 

coefficient is negative. Hence we conclude that if a curve i< 

convex toward the axis of abscissas, the ordinate and second 

differential coefficient will have the same sign / or, the product 

d 2 y . 
y-72 1S positive. 

108. To determine when a curve is concave toward the axis 

of abscissas. 

Let PPT" be a curve concave to- 
ward the axis of abscissas and above 
it, and let x and y be the co-ordi- 
nates of the point P. Let x be in- 
creased by any arbitrary increment 
PR', which we will represent by A, 
and take R'R" also equal to h. Draw 
the orclinates RT', K"P" ; draw the 
secant line PP', and produce it to B ; 
join P' and P", and draw PD,P'D' 
parallel to AR. We shall then have 
EP=/(«)=y; 



I 

/ 


F 


^ 


B 

TV 


y 


D 










R 


R' 


R" 


\ 




d 




p 


X 






X 




d' 




p> 


S 


*L 



DF=BT'-BP=g/ 

BP'=DT"-DF=^+ etc 

| 

r of this equati 

.all that the gqna- 

tram ; ti 

will itive, while the ordinate 

If the cur' -,w the axis of abscissas, we shall have 

where tl tial coefficient fa hile the 

live. I hide that {/' a i 

mhZZ Aaw tipnft on the pro- 

duct y- j/dtiYe. 

A car >lently concave or convex toward the bottom 

of the p r positive. To de- 

ine the "ty or convexity toward any straight line 

whatever, chan^ aking that line the 

d -y 
nev : #. and find the in the ne m. 



132 DIFFERENTIAL CALCULUS. 

Ex.1. It is required to determine whether x 2 + y 2 — r 2 is con- 
vex or concave toward the axis of x. 
Differentiating, we have 

dy x 
dx~ y 
d 2 y x 2 + y 2 r 2 



Also, 



dx 2 y 3 y 3 ' 

d 2 y r 2 



,,2> 



that is, y^ 

which is negative. Hence the curve is always concave to- 
ward the axis of x. 

Ex. 2. It is required to determine whether a 2 y 2 + b 2 x 2 — a 2 b 2 
is convex or concave toward the axis of x. 

Ex. 3. It is required to determine whether y = 5 + 3(x + 2) 2 is 
convex or concave toward the axis of x. Ans. Convex. 

Ex.4. It is required to determine whether y = 8x + 6x 2 — 2x 3 
is convex or concave toward the axis of x. 

Ans. From x=— oo to x— — 1, convex; 
" x——\ to x—0 y concave; 
" x=0 to a?=l, convex; 
" x — 1 to x — 4, concave; 
" x—4c to # = oo, convex. 
Ex. 5. It is required to determine whether y=x*—12x 3 + 
48a? 2 — 64# is convex or concave toward the axis of x. 

Ans. From x~— oo to x = 0, convex; 
" x — to x = 2j concave; 
" x — 2 to x= oo, convex. 

SINGULAR POINTS OF CURVES. 

109. A singular point of a curve is one which is character- 
ized by some peculiarity not belonging to other points of the 
curve. The most remarkable of these are, 1st, points of max- 
ima and minima ordinates; 2d, points of inflection; 3d, mul- 
tiple points; 4th, cusps; 5th, isolated or conjugate points; 
6th, points of stopping. 

Points of maxima and minima ordinates have already been 



SINGULAR POINTS OF CURVES. 133 

considered, Arts. 79 to SS. They differ from other singular 
points in that they depend upon the position of the axes, and 
change by transformation of co-ordinates. 

110. Points of inflection. A point of inflection is a point 
at which, as x increases, a curve changes from convexity to 
concavity toward the bottom of the page, or conversely. Such 
a point is also characterized by the fact that the curve cuts its 
tangent at that point. 

111. To determine points of inflection. When the curve is 
convex toward the bottom of the page, the second differential 
coefficient is positive ; but when the curve is concave, it is neg- 
ative. Hence at a point of inflection the second differential 
coefficient must change its sign. Therefore, between the posi- 
tive and negative values there must be one value equal to zero, 
or infinity ; and the roots of the equation 

cPy d 2 y 

dx 2 ~ ' dx 2 ~ ' 
will give the abscissas of the points of inflection. 

But since a varying quantity does not necessarily change 
sign upon passing through zero or infinity, the points thus de- 
termined must be tested to ascertain whether the second dif- 
ferential coefficient does really change its sign. For this pur- 
pose, we may either construct the curve in the neighborhood 
of the point, or we may substitute in the second differential 
coefficient a value of x a little greater, and one a little less than 
the critical value, and observe whether the second differential 
coefficient has contrary signs for these new values of x. 

Ex.1. Determine whether the curve y = a + (x— h) z has a 
point of inflection. 

Differentiating, we find 

and ^=e(x-i). 



134 



DIFFEEENTIAL CALCULUS. 



Placing this equal to zero, we have x = b, which in the equa- 
tion of the curve gives y~a. To determine whether this point 
is a point of inflection, we find that, when x< b, the second dif- 
ferential coefficient is negative ; but when x>b, the second dif- 
ferential co-efficient is positive. Therefore, there is an inflec- 
tion of the curve at the point y=za,x=b. 

On the left of P the curve falls below 
the tangent line TT', while on the right of 
P it runs above TT'. 

Ex. 2. Determine whether the curve 
y = a—(x—b) 3 has a point of inflection. 
Ans. This curve is first convex, and then 
concave toward the axis of ab- 
scissas, and there is an inflection 
at the point x — b. 
Ex. 3. Determine whether the curve 
24?/ = 3a? + 18a? 2 -2# 3 has a point of inflection. 






a 




By differentiation, we find 

d 2 y 



-x 




dx 2 ~ 2 " 

Putting this equal to zero, we 
obtain x = 3, which gives a point 
of inflection. If x be <3, 3— x 
is positive; if x be >3, 3— x is 
negative. 

Ex.4. Determine whether the curve y=ax + bx 2 —cx 3 has a 
point of inflection. 7 

Ans. x=z— gives an inflection. 

Ex. 5. Determine whether the curve y=x+3Gx 2 — 2x 3 — a? 4 
has a point of inflection. 

Ans. There are inflections at x = 2 and x— — 3. 

112. A multiple point is a point through which two or more 
branches of a curve pass. There are two species of multiple 
points : 1st, where two or more branches of a curve intersect ; 






SINGULAR POINTS OF CURVES. 135 

2d, where two or more brandies are tangent to each other. 
The latter are sometimes called points of osculation. 

113. To determine multiple points. If the equation of a 
curve be of the nth degree, the equation may have n roots 
(Alg., Art. 436) ;. that is, the ordinate y may have n different 
values. Since two or more branches of a curve pass through 
a multiple point, the different values of the ordinate y corre- 
sponding to the abscissa x for such a point become equal to 
each other, while for each value of x a little less or a little 
greater than the former the corresponding values of y are not 
equal. Having found a value of x for which two or more corre- 
sponding values of y are equal, while on each side of this point 
the corresponding values of y are not equal, determine the 

first differential coefficient -*-, and observe whether it admits of 

ctx 

more than one value for this point. If it has two or more real 
values, the curve admits of more than one rectilinear tangent 
at this point ; that is, two or more branches of the curve inter- 
sect, and the number of intersecting branches will be denoted 

nil 

by the number of real roots of -y-. Such a point is a multiple 
point of the first species. -, 

If the equation that gives the value of -j- has two or more 

real and equal roots, different branches of the curve have a 
common tangent at this point, and this will be a multiple point 
of the second species. 

Multiple points must be distinguished from points of a curve 
at which x is a maximum or minimum. Only one branch of 
a curve passes through a point of the latter description. 

Ex.1. Determine whether the curve y — h±.{x—a)-y/x has a 
multiple point. 

We see that for every positive value of x there are generally 
two unequal values of y ; but when a?=0, the two values of y 
become — b; and when x = a, the two values of y become =b. 
On each side of the point x — a y y=b, y has two real values, 



136 



DIFFERENTIAL CALCULUS. 



as we find by substituting for x, in the equation of the curve, 
a+k and a — h successively. This point is therefore a rnul- 
x 




tiple point. 



The value of -¥- is 
ax 

3x — a 
dy 



which f or x = a becomes -j-—± ^/a. Hence 

the tangents to the curve at this point make angles with the 
axis of x whose tangents are + ^/a and — -yja. This is there- 
fore a multiple point of the first species. 

The point x = 0,y = b is not a multiple point, because when 
x is negative, y is imaginary, which shows that the curve does 
not extend to the left of the axis of y. There is therefore but 
one branch of the curve passing through this point, and this is 
simply a point where x is a minimum. 

Ex. 2. Determine whether the curve y 2 — a 2 x 2 —x^ has a 
multiple point. 

Extracting the root of each member, we have 

y=±x(a 2 — a? 2 ) 2 , 
from which we see that for each value of x between the limits 

of +a and —a there are generally 
two values of y with contrary signs. 
Hence the curve has two branches 
which are symmetrical with respect 
to the axis of x. When x—±a^y 
has but one value, viz., 0; but these 
are not multiple points, because the curve does not extend be- 
yond the limits x— +a and x——a. Also, when x — 0^y has 
but one value, viz., 0, while on each side of this point y has two 

values. Here, then, is a multiple point. 




a 2 — 2x 2 



dxi 
The value of / is 
ax 



which, when a?=0, becomes 



(a 2 —x 2 ) 
ax 



Hence one tangent line 



SINGULAR POINTS OF CURVES. 137 

makes an angle with the axis of abscissas whose tangent is + a; 
the other, an angle whose tangent is — a. This is therefore a 
multiple point of the first species. 

Ex. 3. Determine whether the curve ay 2 = bx 2 — x 3 has a mul- 
tiple point. 

Ans. This curve has a multiple point of the first species at 
the origin ; and the tangent of the angle which the 
tangent line makes with the axis of abscissas is 



v 



a 

Ex.4. Determine whether the curve y 2 =x 5 + x 4: has a mul- 
tiple point. 

The two critical points are x—0 and x— — 1. The first dif- 
ferential coefficient is 

5x 2 + 4:X 

which becomes, when # = 0, j~ = 0. The \ 

two branches of the curve are tangents to each other at this 
point, and the origin is a multiple point of the second species. 

114. A cusp is a point of a curve at which two branches 
have a common tangent, and stop at that point. If the two 
branches lie on opposite sides of the common tangent, the cusp 
is said to be of the first species ; if both branches lie on the 
same side of the tangent, the cusp is said to be of the second 
species. 

115. To determine whether a curve has cusps. Since a cusp 
is a variety of multiple points of the second species, the process 
is nearly the same as described in Art. 113 ; but since the 
branches stop at the cusp instead of running through it, the 
values of y on one side of this point are real, and on the other 
side imaginary. To distinguish a cusp from an ordinary mul- 
tiple point, we must therefore compare the ordinate for that 
point with the ordinates of the curve on each side of that point. 



138 DIFFERENTIAL CALCULUS. 

If for any point y and -j- have each but one value, and 
the values of y on one side of this point are imaginary, while 

•r-f has two values with different algebraic signs, this point is 

d 2 y 
a cusp of the first species ; if y^ has two different values with 

the same algebraic sign, the point is a cusp of the second 
species. 

To ascertain the position of all the cusps that may belong to 
any curve whose equation is given, the process above described 
with respect to the axis of X should be repeated with respect 

dx 
to the axis of T, x being regarded as a function of y, and -y- 

being the first differential coefficient. 

Ex.1. Determine whether the curve (y—b) 2 = (x—a) 3 has a 
cusp. 

Extracting the square root, we have 

_3 

y—b±.{x— a) 2 ' 

For x<a, y is imaginary; for x~a,y has but one value, 

viz., b ; while for x>a, y has two real values. Hence the 

curve does not extend to the left of the point x = a,y=b; but 

on the right side of this point the curve 

extends indefinitely, and consists of two 

branches. j o 

The value of -j- is ±-{x-- a)~*~, which be- 

x comes when x—a. Hence the two branch- 
es at P have a common tangent, which is parallel to the axis 
of X, and there is a cusp. To determine the species of cusp, 
we find the value of the second differential coefficient for a 
value of x a little greater than a, as a + h. 

The value of -7— is ± -r(x—a) ¥ . which becomes ±- A — ry when 
dx 2 4: K ! ' 4VA 

the abscissa is a + h. The cusp is therefore of the first species, 

as shown in the annexed diagram, where the upper branch, 

3 
which is convex toward the axis of X, is pointed out by +f~77j 




SINGULAR POINTS OF CURVES. 



139 



and the lower branch, which is concave, is pointed out by 
3 



"4VA" 



z(x 2 — a 2 ) 3 has 




Ex.2. Determine whether the curve (y—b) 2 -- 
a cusp. 

Extracting the square root, we have 

y=b±(x 2 — a 2 )*. 

We find that y is imaginary for any value of x between 
-fa and — a; iovx — dba, 3/ has 
but one value ; while for any 
point to the right of P or to the 
left of P', y has two values. 
The curve has two cusps of x 7 
the first species, one at P, the other at P r . 

Ex. 3. Determine whether the curve (y—x 2 ) 2 =x 5 has a cusp. 

Extracting the square root, we have 

y—x 2 ±x 2 . 
If x is negative, y is imaginary ; for x—0^ y—0 ; for every 
positive value of a?, y has two real values, both of which are 

positive as long as x 2 >x 2 , or x<l ; after which, one is positive 
and the other negative. 

dy „ 5 3. 



which =0 when x~0. It also equals 

i o 5 3. 16 . 

when 2a?±^a? 2 =0, or ai=^. Hence the axis or X is tangent 

to both branches of the curve at the origin, and the tangent to 

the lower branch is parallel to the axis of X at the point M 

1 a 
whose abscissa is ^7. For x — — h the values of y are impos- 




sible. 



(L 2 y 
The values of -j^ are 2 : 



15i 1M1 15 1 

: -rx*. which become 2±:-rnJ 
4 ' 4 



when the abscissa is + A, w T hich are both positive when h is a 
very small magnitude. Hence at the origin there is a union 
of two branches of the curve, both convex toward the axis of X, 
forming a cusp of the second species. 



140 



DIFFERENTIAL CALCULUS. 



Ex. 4. Determine whether the curve (y—x 2 ) 2 = (x—a) 5 has 
a cusp. 

Extracting the square root, we have 

y=x 2 ±(x— d)~*. 
If x<a,y is imaginary; if x=a,y = a 2 ; if x>a, y has two 

5 

real values, both of which are positive as long as x 2 >{x— a) T . 
d 2 y 15, .i , . , m „ 15, 



The values of 



dx 




are 2 ± -j-(# — a) 2 , which become 2 ± -^-A" 2 " 

when the abscissa is a + h, which values 
are both positive when h is a very small 
magnitude. Hence at the point x~a 
there is a cusp of the second species; 
and the tangent of the angle which a 
x tangent line at this point makes with the 
axis of abscissas is 2a. 



116. Definition. A conjugate or isolated point is one whose 
co-ordinates satisfy the equation to the curve while the point 
itself is in general detached from every other in the curve. 

For exam pie, let y 2 = x 3 — x 2 . 

Here the values # = 0, y=0 satisfy the equation to the curve; 
but no branch of the curve passes through the point thus de- 
termined, y being impossible for all negative values of x, and 
also for all positive values less than unity. Hence the origin 
of co-ordinates is a conjugate point in this curve. 

117. To determine whether a point is a conjugate point of 
a curve. Let y—f{x) be the equation of a curve which has a 
conjugate point ; then, if this point is separated from every 
other in the curve, it follows that on each side of this point 
real values of one of the co-ordinates must give imaginary 
values for the other. Let the co-ordinates of the given point 
be a and h (where a and h are supposed to satisfy the equation 
of the curve), and let the abscissa a be increased by a small 
quantity, h; then the corresponding ordinate represented by 
,f(x+h) must be imaginary. 



SINGULAR POINTS OF CURVES. 141 

Now, by Taylor's theorem, Art. 64, 

dy lf d 2 u h 2 d 2 y h 3 

Since y and h are both real, in order th&t f{x+h) may be 

imaginary, it is necessary that one of the differential coeffi- 

du d 2 y . _,_ . . _. . . . 

cients -T-, -7-2, etc., should be imaginary. It, when x = a, either 

of these coefficients becomes imaginary, the corresponding 
point is a conjugate point. In order, therefore, to find the 
conjugate points of a curve, we examine the successive differ- 
ential coefficients, to see if any one of them becomes imaginary 
for a particular value of x which satisfies the equation of the 
curve. 

The labor of forming the higher orders of differential co- 
efficients is often so great that, if the first differential coefficient 
does not become imaginary, it may be more convenient to sub- 
stitute successively a + h and a—h for x in the equation of the 
curve. If both values of y thus found are imaginary, the 
point at which x = a is a conjugate point. 

Ex.1. Determine whether the curve y 2 — x(a + x) 2 has a con- 
jugate point. 

The first differential coefficient -7- = ± 9 . , which, when 

X—— a, becomes —. , an imaginary quantity, although — a 

satisfies the equation of the curve. Hence the point P at 
which x— — a, y — is a conjugate point. 
Extracting the square root of the 
given equation, we have 

y= ± (a+x) -/a?, p 

from which we see that y = when 

# = 0, and also when x——a; but for 

any other value of x less than zero, y is imaginary. For every 

positive value of x, y has two real values, which shows that the 

curve has two branches extending indefinitelv toward the 

right. The point P is therefore entirely isolated, and no 



142 DIFFERENTIAL CALCULUS. 

branch of the curve passes through it, although the co-ordi- 
nates of P satisfy the equation of the curve. 

Ex. 2. Determine whether the curve ay 2 =x 2 (x—b) has a con- 
jugate point. d _ b 

Ans. At the origin -y-= — ==, although the co-ordinates 
(ix y — qJj 

x=0^y — satisfy the equation of the curve. Hence 
the origin is a conjugate point. 

TRACING OF CURVES. 

118. To trace a curve is to discover from its equation its 
general form and leading peculiarities, and its position with 
reference to the assumed axes, so that we may draw a correct 
figure of the curve. 

If it was possible to resolve an equation of any degree, we 
might follow the course of a curve represented by any alge- 
braic equation by methods explained in Analytical Geometry. 
By ascribing to one of the variables different values, both pos- 
itive and negative, we could determine at pleasure any number 
of points of the curve. 

The Differential Calculus enables us to abridge this investi- 
gation, and may be employed even when the equation of the 
curve is of so high a degree that we are unable to obtain a gen- 
eral expression for one of the variables in terms of the other. 

119. To trace a plane curve from its equation referred to 
rectangular axes. It is impossible to give specific directions 
for tracing the curve corresponding to every equation, for the 
methods which are most convenient in certain cases may not 
be the best for other cases. 

1. The equation should, if possible, be solved with reference 
to one of the variables ; that is, it should be put under the form 

y =/(«)• 

2. Observe whether the curve is symmetrical with respect to 
either of the axes; that is, whether x or ?/, or both of them, 
enter into the equation only in even powers. 






TKACING OF CURVES. 143 

3. Observe whether the origin is a centre of the curve ; that 
is, whether the equation remains unaltered when y is changed 
into — y, and x into — x. The centre of a curve is a point so 
situated that all chords of the curve drawn through that point 
are bisected in it. 

4. Observe whether the curve passes through the origin ; 
also, the shape of the curve near the origin. 

5. Determine the points where the curve cuts the two axes, 

dy 
and find -£■ for those points. 

6. Observe for what values of x, y is real, and for what values 
y is imaginary, and conversely. When there are several values 
of ?/, find for what values of x couples of them become imag- 
inary. From the real values of y the number and position of 
the branches may be found. Remember that the degree of 
the equation of an algebraical curve determines the number 
of points in which a straight line may cut the curve. 

7. Determine the limits of the curve in the direction of the 
axes, and ascertain if there are infinite branches by putting x 
or y equal to infinity. 

8. Examine infinite branches for asymptotes, and determine 
them if they exist. 

9. Observe whether for particular points of the curve the 

first differential coefficient of y reduces to 0, oo, -, an imagin- 
ary quantity, etc. 

10. Determine the position of maxima and minima ordi- 
nates, and maxima and minima abscissas. 

11. Find the second differential coefficient of 3/, and deter- 
mine whether the curve is convex or concave, and find the 
points of inflection. 

12. Having determined the number and position of the sin- 
gular points, examine the course of the curve in their imme- 
diate vicinity. 

Ex. 1. Trace the curve whose equation is y= 1 ,. 

Iii order to illustrate the preceding directions, w T e will follow 



144 



DIFFERENTIAL CALCULUS. 



them in succession as far as they are applicable to the present 
example. 

(1) The equation is already reduced to the required form. 

(2) The curve is not symmetrical with respect to either axis. 

(3) The origin is a centre of the curve. 

(4) and (5) The curve cuts the axis of x at the origin and at 
infinity, and the axis of y at the origin only. Near the origin 
(since x 2 is small), y is nearly equal to a?, and numerically less 
than it. Therefore, near the origin, the curve lies between the 
axis of x and the straight line y~x. 

(6) and (7) For each value of x there is one, and only one, 
value of y, and this has the same sign as x. There is therefore 
one branch extending from x~— oo to a?=+oo, and lying in 
the first and third quadrants. 

Let AX and AY be the axes of x and y respectively, and A 

the origin of co-ordinates. 
Draw the line TAT' whose 
equation is y=x. In the 
first quadrant the curve 
will lie below AT, and it 
will touch AT at the or- 
igin. In the third quadrant the curve w T ill lie above AT 7 . 

(8), (9), and (10) By differentiation, we find ^= / 1+ ^a - 
dy 





M. X 



Now -j- — when 1 
ax 



-x 



2 = 0, and also when l-\-x 2 —oo. 



When 1 — x 2 = 0, or #=±1, y=±--J-, and we have a maxi- 
mum ordinate at C, and a minimum ordinate at c. 

When l + x 2 — oo^y=0 y and the infinite branch has the axis 



of x for its asymptote. 



(11) and (12) We find, also, ^=-"7^: 



d 2 ;/ 2x(3-x 2 ) 
v 2 ) 3 ' 



This expression changes sign when 2x(3— x 2 ) changes sign; 
that is, when # = 0, and when x=±V3. Hence there are 
points of inflection at the origin, and at P and p. From P to p 
the curve is concave to the axis of x ; elsewhere it is convex. 

Ex. 2. Trace the curve whose equation is a 3 y = x(x—b) 3 . 



TRACING OF CURVES. 



145 



(2) and (3) The curve is not symmetrical, nor is the origin 
the centre of the curve. 

(4) Developing the equation, we have 

a 3 y = x± -Sbx 3 + 3b 2 x 2 - h 3 x. 

Since there is no constant term, the curve passes through the 
origin. When x is small, the equation becomes very nearly 
a 3 y=—b 3 x, and this straight line is a tangent to the curve at 
the origin. 

When x is large, the lower powers of x are not important, 
and the equation becomes very nearly a 3 y — x^. 

(5), (6), and (7) For each value of x there is one, and only 
one, value of y, and therefore there is a single branch from 
X— —oo to x= +00. 

The value of y is plus when x is minus, and also when x>b. 
It is minus from x=0 to x — b. 

(9) and (10) Differentiating, we have 
dy (x—b) 2 (4:X—b) 



dx~ 

d 2 y 
dx 2 ~ 

d 3 y 



a° 

6(x-b)(2x-b) 
a 3 ' 

2±x- 18b 



dx 3 



a° 



sis)! 7)3 

At the origin ^— — §-, and the line a 3 y=—b 3 x is tangent 



dx' 

to the curve, as shown above, 
dy 



dx~ 



when x=b, and when 4cX=b. When x 



and we must examine the value of 

and hence, Art. 86, x — b does M 

not give either maximum or 

minimum values of y. The 

b d 2 y 

second value, x — j, makes -r-^ 

positive, and hence we have a 
minimum value of y at D. 



This reduces to 




146 



DIFFERENTIAL CALCULUS. 



d 2 y b 

(11) The sign of -j4 changes from plus to minus at x-=-, 

and from minus to plus at x=b. Hence, from F to B the 
curve is concave downward, and elsewhere it is convex; and 
F and B are points of inflection. 

Ex. 3. Trace the curve y 3 =a 3 —x 3 . 

When x—0, y=a; take AB=#, and the curve passes 
through B. When y=0, x=a; take AC = #, and the curve 
passes through C. When x is infinite and positive, y is infinite 
and negative ; therefore the arc CR is below AX, and extends 
indefinitely. When x is infinite and negative, y is infinite and 
positive ; therefore the branch BQ extends indefinitely. 



dy 
dx~ 



x" 



25 
y3 o?"\ " 



which becomes infinite when x—a. Hence 



(a 3 — x 3 y 

at C the curve cuts the axis of X at right angles. 

d 2 y 2a 3 x A d 3 y 2a 3 (4:X 3 + a 3 ) 

(a 3 -x 3 f 

d *y a a d *y 



2a 3 x t d 3 y 

When x 




is not 



°> ^=°» and a? 

zero. Hence there is no maximum or 
minimum at B. 
When x is — , 



d 2 y . 

CLX -f 

i . i <* y . 

when x is + and <a, -j-$ is — ; 



dx 2 
d 2 y 



when x is + and >a, -jrj is +. 
Hence the arc QB is convex, BC concave, and CR convex to 
bottom of the page. 



When # = or &, -^f changes sign, and 



therefore at B and C are points of inflection. 
The equation to the asymptote EF is y= — x. 

Ex.4. Trace the curve V — ~-h; — =*. 

For each value of x there is one, and only one, value of y. 
The curve is not symmetrical to either axis, but the origin is 



TRACING OF CURVES. 



147 



x 2 -\-\ 
the centre. Since 2 ^ is greater than 1 when x is greater 

than 1, y is in that case greater than x. 

When #=0, y=0; therefore the curve passes through A. 
When x<lj y is negative, and the curve lies below AX. When 
8=1, y is infinite. Draw the ordi- 
nate BD, and produce it indefinite- 
ly ; it will be an asymptote to the 
branch AQ. 

When x>l, y is positive; when 
x— 1 is very small, y is very great; 
and when x — 1, y is infinite. There- 
fore BD produced is an asymptote, 
and the branch RFS is infinite. 

When x is negative and less than 
1, y is positive, and the curve lies 
above X'AX. When a?= — 1, ?/ is infinite. Take Ai= — 1, 
and the ordinate 6<i is an asymptote, the branch Aq being in- 
finite. When a?— 1 is very small, y is very great; therefore be 
produced is an asymptote to the branch rfs. 

dy x 2 (x 2 -4:)-l '_ ., 

^"^ — r^ — tt^ — , which reduces to —1 when x=0: hence 

ax (x 2 —!) 2 ' 

at the origin the curve cuts the axis of X at an angle of 135°. 

When ^=0, x= ± V2 + <y/5 ; hence, if AE= Ae= V2 + V&, 
the ordinates EF, <g/*will be minima. 




d 2 y 4:x(x 2 + S) 



which is — when x is + and <1, or — and 



dx 2 " {x'-lf > 

>1. It is also + when x is + and >1, or — and <1; there- 
fore every part of the curve is convex toward the axis 
of X. 

d 3 y 12( x 4 +6x 2 + l) 

dx 3 ~ 



When 05=0, g=0, but g 
point of inflexion. 



(x 2 -iy ■ 

-12 ; hence at A there is a 



148 



DIFFERENTIAL CAXCULUS. 



The equation to the asymptote TAt is y—x. 

x Ex. 5. Trace the curve y 3 =a 2 x—x 3 . 







dy 



a° 



• 3« 2 



dx' 
d 2 y 



3(a 2 x-x 3 f 
2a 2 (3x 2 +a 2 ) 



dx 2 



§{a 2 x-x 3 f 



The equation to the asymptote is 



Ex. 6. Trace the curve (a—x)y 2 =(a+x)x 2 



dy a 2 +ax—x 2 /a+x 
dx~ a?—x z V a—x> 



<py 
dx' 



= =fc- 



a 2 (2a+x) 



3* 



(a—x)(a 2 —x 2 y 
The equation to the asymptote is x~a. 





Ex. 7. Trace the curve y 3 =ax 2 —x 3 . 

There is a cusp at A, and a point 

x of inflection at B, and the asymptote 

makes with the axis of X an angle 

of 135°. 



Ex. 8. Trace the curve y 2 — 



x 6 



x—a 




TRACING OF CURVES. 



149 



^ ^ rr, , X(x+1) 2 

Ex. 9. Trace the curve y~- ( r^r. 

9 (x— l) 2 

y=x+4: is the equation to one asymp- 
tote. 

dy x 3 —3x 2 —5x—l 
dx= (x-lf ~5 

d 2 y 8(2a?+l) 
cfc 2 ~ (#-l) 4 * 





Ex.10. Trace the curve y 2 (2a—x)=x 3 

dy (3a—x)\/x 
dx 



3 



x=2a is tlie equation to the asymptote. 



150 DIFFERENTIAL CALCULUS. 




CHAPTER VII. 

DIFFERENTIAL COEFFICIENT OF AN ARC, AREA, ETC. 

120. The limiting ratio of the length of an indefinitely 
small arc to its chord is one of equality. Let ADB be an 

arc of any curve, and let it be taken so 

small that it is always concave toward 

its chord AB. At the extremities of 

this arc let the tangents AC, BC be 

drawn. The arc ADB is greater than 

the chord c, but less than the sum of the two tangents a and b 

(Chauvenet's Geom., Bk. V., Prop. 12). 

By Trigonometry, Art. 49, 

<z_sin.A b sin.B 

c~~sin. C c~ sin.C 

. a + b sin.A + sin.B sin. A + sin.B 

hence c - ^ Q = gin(A+B) . 

By Trigonometry, Art. 76, 

sin. A + sin. B cos. |(A — B) 
sin.(A + B) = cos.|(A + B) ; 

a+b cos.^(A— B) 

hence = 1/A , t^ . 

c cos. £( A + B) 

Let now the points A and B approach each other, and the 
arc ADB diminish continually. The angles A and B will de- 
crease ; and the arc may be taken so small that the angles A 
and B shall be less than any assigned quantity ; therefore 
A— B and A+B both approach continually to zero; and 
cos. -g-(A— B) and cos. ^(A + B) approach to unity, which is 
their common limit. Hence the limiting ratio of a+b to c is 
a ratio of equality ; and since the arc ADB can not be greater 
than a+b , nor less than c, much more is the limiting ratio of 
the arc to its chord a ratio of equality. 



DIFFERENTIAL COEFFICIENT OF AN ARC, AREA, ETC. 151 



121. The differential of an arc of any plane curve referred 
to rectangular co-ordinates is equal to the square root of the 
sum of the squares of the differentials of the co-ordinates. 

Let AE and EP be the co-ordinates 
x and y of any curve. If now we give to 
x any arbitrary increment A, and make 
BB'=A, the value of y will become equal 
to P'B', which we will represent by y'. 
If we draw PD parallel to the axis AB', 
we shall have 

the chord PP' = VPD 2 + P'D 2 = a/A 2 + PT) 2 . 

dv d 2 v h 2 

But PT) = y' — y = -^h + -J^~2 + other terms involving high- 
er powers of h (Art. 64). 

Substituting this value of P'D in the expression for the 
chord, we have 




PP 



■v 



/r- 



dy 2 h 2 

-kr + etc - 



=A v 1+ %+ etc - 



pp' f~W 

-A--V 1 +^+ etc., 



Therefore 

which expresses the ratio of the chord PP' to the increment 
of the variable, and we must find the limit of this ratio by 
making the increment equal to zero, Art. 18. 

But, by Art. 120, the limit of t = the limit of — ^ — ; 

and the terms omitted in the second member of the equation 
contain h, and therefore disappear. Hence, representing the 
length of the arc by L, we have 

dh 

dx~ 
or, multiplying by dx. 



Vi+g; 



dL=Vdx 2 + dy 2 . 



152 DIFFEEENTIAL CALCULUS. 

Ex. 1. Find the differential of the arc of a circle. 



From the equation x 2 + y 2 = r 2 , 



we find xdx + ydy=0; whence dy—- 



V 



Hence dJj = V dx 2 + — ^— = — Vx 2 + y 2 . 

v y 2 y -rv 



But Vx 2 +y 2 =r y and y= Vr 2 —x 2 . 
Hence dh — - 



V?' 2 —x 2 
Ex. 2. Find the differential of the arc of a parabola. 



Ans. dL = \/^±^.dx. 
v x 



122. The differential of any plane area included between a 
curve and the axis of x, and referred to rectangular co-ordi- 
nates, is equal to the ordinate into the differential of the ab- 
scissa. Let APM be any plane area lim- 
ited by a curve and the axis of x ; and let 
x represent the abscissa AM, and y the 
corresponding ordinate PM. If we give 

Jl~ ~~l{[i ]$ to x an increment A, and make MN=A, 

the value of y will become QN, which we will represent by y '. 
Through Q draw QE parallel to AN, meeting MP produced 
in E, and represent the area of APM by A. We shall then 
have incr. x = A, incr. y = DQ, and incr. A = area MPQN. Also, 
area MEQN_EM.MN_EM_2/ + incr.y incr.?/ 

area MPDN~PM.MN = PM=~ ~Y~ ~ = ~~V~~' 
Now, as h approaches zero, incr. y also approaches zero ; 
hence, if we take the limiting values of both sides of this 
equality, we shall have 

.. . c area MEQN _ 

limit 01 MPnAT - 1 ' 

area MPDN 

But area MPQN is intermediate between the area MEQN 

and the area MPDN. Hence 

„ . * area MPQN H 

limit of MpmT = l. 

area MrDJN 



DIFFERENTIAL COEFFICIENT OF AN ARC, AREA, ETC. 153 

* 

Now area MPQN=incr. A, and area MPDN=y. A. 

Hence limit of 7— = 1 ; 

y.h ' 

dJA 

ydx~ 

Therefore -r-—y^ or dA=ydx. 

In a similar manner, it may be proved that the differential 
of the area included between the curve and the axis of y is 

dA=xdy. 
Ex. 1. Find the differential of the area of a circular segment. 
When the origin is at the centre, the equation is 

yi _ r 2 — x 2 9 

Hence dA — ydx — dx Vr 2 — x 2 . 

When the origin is on the circumference, the equation is 
y 2 = 2rx— x 2 . 

Hence dA = dx V%rx — x 2 . 

Ex. 2. Find the differential of the area of an ellipse. 



Ans. dA = dx- Va 2 — x 2 . 
a 

123. The differential of the surface of a solid of revolution 
is equal to the circumference of a circle perpendicular to the 
axis multiplied by the differential of the arc of the generating 
curve. 

Let APQ be a curve which, by revolving about the axis of a?, 
generates a solid of revolution. Let AM=#, 
and PM=2/. If we give to x an increment 
A=MN, the value of y will become QN. 

Let S = the surface generated by the revo- 
lution of the arc AP( = L) round the axis 
AN, and incr. S = the surface generated by 
the revolution of the arc PQ or incr. L, due to the increment 
MN given to x. 

Draw PD and QE each equal to PQ, and each parallel to 

G2 



E Q^-^ 


P^" 


n 


/^ 







154 DIFFERENTIAL CALCULUS. 

AN". If PD revolve about AN, it will generate a cylinder 

whose surface is 27rPM. PD. If QE revolve about AN, it will 

generate a cylinder whose surface is 27rQN . QE. Hence 

surf, gen, by QE 2?rQN . QE QN y + incr. y incr. y 

surf. gen. by PD~2^P]0^~PM~~ ~^ ~V~' 

mi r • -,. .^ 1 « surf . gen. by QE . 

Ihererore limiting; value of — 7-^ , t»t^ = 1. 

& suri. gen. by PD 

But the surface generated by the arc PQ is intermediate 
between the surface generated by QE and the surface gen- 
erated by PD ; hence 

,. ... , r surf . 2;en. by PQ • 

limiting value of — t^ 2 . pn =l; that is, 

to suit. gen. by ID 

limiting value of -5 =— : — T =1« 

& 2-n-y mcr. L y 

.... _ „ incr. S ^ 

or limiting value 01 -. — — y-=27ry. 

Hence -Tj — 2iry 7 or dS = 27rydL. 

Substituting for dL its value, Art. 121, we have 

dS = 2TryVdx 2 +dy 2 . 
If the curve revolve about the axis of y, we shall have 

dS = 27rxVdx 2 +dy 2 . 
Ex. 1. Find the differential of the surface generated by the 
revolution of a parabola about its axis. 

The equation of the parabola is y 2 =:&ax; whence 

dx=-^. 

Therefore dS = 2 7 ry\/^^+dy 2 = 7 ^Vy 2 + 4:a 2 . 

Ex. 2. Find the differential of the surface of a sphere. 

Arts. dS = 27rrdx. 

124. The differential of the volume of a solid of revolution 
is equal to the area of a circle perpendicular to the axis mul- 
tiplied by the differential of the abscissa of the generating 



DIFFERENTIAL COEFFICIENT OF AN ARC, AREA, ETC. 155 




curve. If the surface APM revolve around 
the axis of x, it will generate a solid of rev- 
olution. Put AM = a?, and PM = y. If we 
give to x an increment A = MN, the value 
of y will become QN. 

Let V= the volume of the solid generated 
by the revolution of APM around the axis 
AM. Then incr. x~h^ incr. ?/=:DQ, and 
incr. V = the solid generated by the revolution of MPQN. Now 
the rectangles PN and EN by their revolution will generate 
cylinders whose volumes may be found by Geom., Bk. X., 
Prop. 2. Hence 

vol. cyl. EQgg_ 7rEM 2 .MN EM 2 (y+incr.y) 2 __/ incrjA 2 
vol.cyl.PD^~7rPM 2 .MN~PM 2 ~ tf "\ + y /' 

Now, as h approaches zero, incr. y also approaches zero ; hence, 
if we take the limiting values of both sides of this equality, 
we shall have # vol L EQ 

limitof vol.cyl.PD^ ^ L 

But the solid TQqj) is alwa) 7 s intermediate between the 
two cylinders ; hence vq1 pQ 

limit of vol ; FDdp . 

Now volume TQqj) = increment V, and volume PDdp=Try 2 h. 



=1. 



Therefore 



limit of - T^r-=lj 



iry 2 h 



dV 



or 



iry 2 dx 



=i; 



that is, 



dx 



=7r?/ 2 , or dV=wy 2 dx. 



Ex. 1. Find the differential of the volume of a paraboloid of 
revolution. 

The equation of the parabola is y 2 — ^.ax; whence 

ydy 



dx— y 



2a' 



156 DIFFERENTIAL CALCULUS. 

Therefore dV= L . — 4ciraxdx. 

2a 

Ex. 2. Find the differential of the volume of a prolate 

Ans. dV=—(a 2 — x 2 )dx. 

TOLAR CURVES. 

125. The differential of an arc of a curve referred to polar 
co-ordinates is equal to the square root of the sum of the 
squares of the differential of the radius vector ', and of the 
product of the radius vector by the differential of the varia- 
ble angle. 

Let P be the pole of the curve AM1ST ; let the radius vector 
PM be represented by r, and the vari- 
able angle MPX by 9. Let the varia- 
ble angle be increased by MPN, which 
we will represent by h ; the value of r 
will become equal to PN, which we 
will represent by r\ Draw the chord 
MN. Then in the triangle PMN (Trig., Art. 78) we shall have 
MN 2 = PM 2 + PN 2 -2PM.PN cos. MPN. 

But cos.MPN = cos.A=l-2 sin. 2 ^ (Trig., Art. 74). 

Hence MN 2 =7*HV 2 — 27r'/l— 2 sin. 2 ^); 

=z(r f —r) 2 +4^rr f sin. 2 ^. 
Dividing by A 2 , we have 

MN 2 (r'-r) 2 




/sin. zk\ 



h 2 ~ h 2 

v \h 

It is required now to find the limiting value of this ratio, 

as the point N is supposed to move toward M until it coincides 

with it. 

mi .. . ,MN 2 , ^L 2 
The limit of —j^- is -^ ; 



POLAR CURVES. 157 

(rp f rv*\2 dr 2 

the limit of ^ is ^; 



/ ■ ^\ 2 
and the limit of y/( sm -2 ) is r 2 

V U ) 



dL 2 dr 2 



Therefore -^ = ^- 2 + r>; or dL= Vdr 2 + r 2 d0 2 . 

Ex.1. Find the differential of the arc of the logarithmic 
spiral. 

The equation of the logarithmic spiral is 0=log. r. 

Consequently dQ = . 

Hence by substitution we find 



dL=drVl + M 2 . 
Ex. 2. Find the differential of the arc of the spiral of Archi- 
medes. A „ T dr 



Ans. dJj— — Va 2 + r 2 . 



a 



126. The differential of the area included between the initial 
line of a polar curve, the arc of the curve, and a variable 
radius vector, is equal to the differential of the variable angle 
nmdtijplied by half the square of the radius vector. 

Retaining the figure and notation of the last article, let 
the variable angle receive the increment MPN ; then will the 
area APM receive the increment MPN. Now the triangle 
PMN = ^PM . PN sin. MPN = \rr' sin. h. Hence 

triangle PMN , ,sin.A 

It is now required to find the limiting value of this ratio, 
as the point N moves toward M until it coincides with it. The 

,. . , triangle PMN , .. - „ curv. area PMN ,. , A 
limit or ^-j =the limit 01 j = limit 

of \rr' 55p = limit of \rr' (Art. 43). 



158 



DIFFERENTIAL CALCULUS. 



Hence, representing the curvilinear area by A, we have 

dA. r 2 
dO = 2> 

or dA=— —. 

Ex. 1. Find the differential of the area of the spiral of 

Archimedes. . ,. r 2 dr 

Arts. dA = -rr-. 

2a 

Ex. 2. Find the differential of the area of the logarithmic 



spiral. 



Ans. dA= 



^ILrdr 




CONTACT, CURVATURE, ETC. 159 



CHAPTER VIII. 

CONTACT. CURVATURE. EVOLUTES AND INVOLUTES. THE CYCLOID. 

127. Different orders of contact. Two or more curves are 
said to touch each other when they have a common point, and 
a common tangent at that point. Thus j^_ 
the curves mn,pq, rs, having the common w- 
point P, and the common tangent AB, are * / 
said to touch each other at P. When 

, three curves touch each other at the same point P, the curve 
j9£, which passes between the two others in the immediate 
neighborhood of P, is regarded as having a closer contact with 
the curve mn than the curve rs has. Geometers thus distin- 
guish different orders of contact, and these orders of contact 
are determined by the differential coefficients of the equation 
of the curve. 

128. How different orders of contact are distinguished. 
Let there be two curves whose equations are y—f(x) and 
y r — Y{x!) ; let them be referred to the same rectangular axes; 
and let their abscissas be increased by a common increment h. 
We shall then have (Art. 64) 

dy 1 d 2 y h 2 d 3 y h 3 

and F(x'+h)=y' + ^h+^- 2 -^ +J- 3 + etc. (2) 

Then, if we suppose the curves to have 
the common point P, x will be equal to x' 
and y to y' for P. If the co-ordinates of 
this point be substituted in the preceding 
equations, f(x + h) will represent the or- 
dinate P'E/, and F(x' + h) will represent 

dv d 2 v 
the ordinate P"R'. Also, -p -y^, etc., 




160 DIFFERENTIAL CALCULUS. 

will represent the successive differential coefficients of the or- 

dy f d 2 y f 
dinate of the first curve, taken at the point P, and -r-„ y-7^, etc., 

the corresponding values for the second curve. 

Since the first differential coefficient represents the tangent 
of the angle which a tangent line makes with the axis of ab- 
scissas, if we suppose not only that x=x f and y=y', but also 

. dy dy f _ .__ . _ 

that -j-=-j-7i the two curves will nave a common tangent at Jr. 
ax dx J & 

This is called a contact of the first order. 

TJ , , - t . , dy dy ! d 2 y d 2 y f 

If we have at the same time y = y , ^=^?, and ~^=^ 

the curves are said to have a contact of the second order. 
TP , dy dy r d 2 y d 2 y' . d 3 y d 2 y' , 

If y=y ' d-x=M> d&=M-» and as^s* the curves are 

said to have a contact of the third order, and so on. 

Now if all the terms of equation (1) are equal to the corre- 
sponding terms of equation (2), the two curves will coincide 
throughout, or be identical; and the greater the number of 
terms in the two developments which are equal, the closer will 
be the contact of the curves at P. 

129. Contact of the first order. In order that a curve may 
have a contact of the nth order with a given curve, it appears 
from Art. 128 that n + 1 equations must be satisfied. Hence, 
if the equation to a species of curves contains n+1 constants, 
we may, by assigning proper values to these constants, find the 
particular curve of the species that has a contact of the nth 
order with a given curve at a given point. 

For example, the equation to a straight line is of the form 
y — mx + c; and since there are two constants m and c,we may, 
by properly determining them, find the straight line which has 
a contact of the first order with a given curve at a given point. 

Let the equation of the proposed curve be y 2 = 16x, and let 
y'=mx' + cbe the equation of the straight line; then we have 

H 3 d2 y f 7, 

— =m,and^=0. 






CONTACT, CURVATURE, ETC. lGl 

At a point of contact of the first order we must have 

. , _ dy dy f 

whence -j-=m, and c — y—mx. 

For the point at which 05=1, y=£, -j- — m—^-—2^ and 
c=y — 2x=2. y 

Hence y=2x + 2 is the equation of the straight line which 
touches the curve y 2 =zl6x at the point x — l^y — ^ and this 
contact is of the first order. In general, a straight line can 
have no higher order of contact than the first. 

130. Contact of the second order. The general equation to 
a circle contains three constants, and hence a circle may be 
found which has a contact of the second order with a given 
curve at a given point. 

Let it be required to find the equation of the circle which 
has a contact of the second order with the curve y 2 = lQx at 
.the point x = l, y=£. 

Differentiating the given equation twice, we find 

dy 8 . . ' 

•r-;=-, which at the given point becomes —2\ 

d 2 y 64 1 . , , . 

33 = — —3 , which at the given point becomes =— 1. 

The most general equation of the circle is 

(x — a) 2 + (y — b) 2 — r 2 . 
Differentiating twice, we have 

(x-a)dx+(y-b)dy=0,whence ^g= — r^j/ 

d 2/ u dx 2 

da*+dy* + (y-b)d 2 y=0, whence ^= — — j- 

Hence for the given point we have the three equations 
(l-a) 2 +(4-J) 2 =r 2 ; 
1-a 



162 DIFFEEENTIAL CALCULUS. 

whence 0=11, b=— 1, ^=5^5. 

Hence the equation of the required circle is (#— ll) 2 + (2/ + l) 2 
= 125. In general, a circle can have no higher order of con- 
tact than the second. 

131. Oscidating circle determined. We may determine in 
a similar manner the general equation of the circle which has 
a contact of the second order with any given curve at a given 
point. Let the equation of the circle be (x—a) 2 + (y—b) 2 = r 2 , 
where a and b are the co-ordinates of the centre, and r the 
radius of the circle. 

Differentiating the equation of the circle, we have 

whence r 2 =(y-i) 2 +^-b) 2 ^=(y-b) 2 ^^); 

2 _ / dx 2 +dy 2 \ 2 / dx 2 + dy 2 \ _ (dx 2 +dy 2 ) 3 
or r- \d a yM<& a / (dxd 2 yf ' 

(dx 2 + dy 2 f 
whence y=± <fe4 > (3) 

which is a general expression for the radius of the circle 
which has a contact of the second order with any given 
curve. 

Curves which have a contact higher than the first order are 
said to osculate; and a circle which has a contact of the 
second order is called an osculating circle. 

132. Curvature of different curves compared. The curva- 
ture of a plane curve is its deviation from a straight line ; and 
of two curves, that which departs most rapidly from its tan- 



163 





gent is said to have the greatest curvature. 
Thus, of the two curves AC, AD, having 
the common tangent AB, the latter de- 
parts most rapidly from the tangent, and 
is said to have the greatest curvature. 

The curvature of the circumference of a circle is the same 
at all of its points, and also in all circumferences described 
with equal radii; but of two different 
circumferences, ACE, ADF, that one 
which has the least radius departs most 
rapidly from the tangent line. The 
curvature of a circle is therefore usu- 
ally measured by the reciprocal of its 
radius ; that is, in two different circles 
the curvature varies inversely as their radii. 

Since the osculating circle has a closer contact with a given 
curve than any other circle, its curvature is regarded as the 
same as that of the curve at the point of osculation. Hence, 
to compare the curvature at different points of a curve, we 
have only to compare the curvatures of the osculating circles 
drawn at those points. Thus, if r is 
the radius of the osculating circle at 
P, and r' that at P', we shall have 

curv. at P : curv. at P': : - : — : 

that is, the curvature at different 
points of a curve varies inversely as the radius of the oscu- 
lating circle. 

If an osculating- circle be drawn to an ellipse at A, the 
vertex of its major axis, it will lie wholly within the ellipse ; 
if an osculating circle be drawn at 
B, the extremity of the minor axis, 
its circumference will pass without 
the ellipse ; at any other point, as 
P, it will cut the ellipse, as will be 
seen hereafter. 





164 DIFFERENTIAL CALCULUS. 

133. Radius of curvature. The radius of the osculating 
circle at a given point of a curve is called the radius of curva- 
ture at that point, and the centre of the osculating circle is the 
centre of curvature. A general expression for this radius is 
given in Art. 131. To find the radius of curvature for any- 
particular curve, we must differentiate the equation of the 
curve twice, and substitute the values of dx, dy, and d 2 y in the 
general expression for r. If the radius of curvature for a par- 
ticular point of the curve is required, we must substitute for x 
and y the co-ordinates of the given point. 

Since in determining the curvature of a curve, we require 
only the absolute length of the radius of curvature, we may 
employ either the + or — sign of the formula. 

134. To find the radius of curvature at any point of a 
conic section. The general equation to a conic section (Anal. 
Geom., Art. 234) is y 2 — mx + nx 2 ; 

, .. (?n+2?ix)dx 
whence dy=- — — , 



2y 



? 



j 7 o 7 o [4:y 2 + (m + 2nx) 2 ]dx 2 
and dx 2 + dy 2 = ± 2 — — . 

Also d 2 v - 2nydx2 ~ ( m + 2nx ) dxd V 

[4:?iy 2 — (m + 2nxf\ dx 2 m 2 dx 2 

(dx 2 + dv 2 )^ 
Substituting these values in the equation r= , , j ^ 

obtain 4 3 ^ 

r= — {mx+nx 2 +^(m + 2?ix) 2 } 1 *. (1) 

9Y1 2 

Iii this expression, -j- is the square of half the latus rectum, 

and the other factor is the cube of that part of the normal in- 
tercepted between the curve and the axis of x (Art. 95) ; that 
is, the radius of curvature at any point of a conic section is 
equal to the cube of the normal, divided by the square of half 



CONTACT, CURVATURE, ETC. 165 

the latus rectum ; and the radii at different points of the same 
conic section are to each other as the cubes of the correspond- 
ing normals. 

If in -equation (1) we make x = 0, we have r — -^\ that is, 

the radius of curvature at the vertex of the major axis of any 
conic section is equal to half the latus rectum, which for the 

ellipse and hyperbola is 



a 
If it is required to find the radius of curvature at the vertex 

of the minor axis of an ellipse, we make m = — , 7i= — —# and 
x=a, which gives, after reduction, 

a 2 

that is, the radius of curvature at the vertex of the minor axis 
of an ellipse is equal to one half the parameter of that axis. 

In the case of the parabola, n=zO, and the general value of 
the radius of curvature becomes 3 

(m 2 + kmx^ 
~~ 2m 2 

Ex. 1. What is the radius of curvature for a point on a par- 
abola whose abscissa is 9 and ordinate 6, the origin being at the 
vertex ? What are the co-ordinates of the centre of curvature ? 
Construct the parabola, with the osculating circle in position. 

Ex. 2. Find the radius of curvature at the vertices of the 
axes of the ellipse whose axes are 10 and 6. 

Ex. 3. Find the radius of curvature for a point of the same 
ellipse at which x — 2. Find also the centre of curvature, and 
construct the osculating circle. 

Ex. 4. Find the radius of curvature for the logarithmic 

curve a? = log. y. , -, Jt 

& x (m 2 + y 2 ) 2 

Ans. r=- "-. 

my 

Ex. 5. Find the radius of curvature at any point in the 

cubical parabola y 3 =ax. /n , , 9X -| 

Ans. r=^^. 
6a 2 y 



166 DIFFEKENTIAL CALCULUS. 

Ex. 6. Find the radius of curvature in the equilateral hy- 
perbola, referred to its asymptotes, xy=m 2 . (#2_|_ ? .2\t 

Am.r=—^r-. 

Ex. 7. Find the radius of curvature for the curve y^—a?x. 

(16y 6 + a 6 f 
Ans. v- 12a y . 

EVOLUTES AND INVOLUTES. 

135. Evolute and involute defined. If osculating circles 
be drawn at every point of a curve, the curve line traced 
through their centres is called the evolute of that curve. Any 
curve, when considered with respect to its evolute, is called an 
involute, for reasons which will be hereafter explained. 

Let pp'p/'P'" be a plane curve, and let the centre of curva- 
ture for the point P be C ; then, as the 
point moves along the curve to P', P", 
etc., the centre of curvature will describe 
another curve, 0, C, C /; , etc. The curve 
CC //7 thus described is called the evolute 
of the curve PP'", and PP /;/ is the invo- 
lute of CC". Any number of points of 
the evolute may be determined by draw- 
ing normals at different points of the in- 
volute, and on each of these normals laying off the correspond- 
ing value of the radius of curvature as given in Art. 131. 

136. Given the equation of a plane curve, to find the equa- 
tion of its evolute. The equation of the evolute is the equa- 
tion which expresses the relation between the centres of all the 
osculating circles of the involute. 

The general equation of the circle is 

(x-ay + {y-bf=r\ (1) 

where a and b denote the co-ordinates of the centre of the 
circle. To determine the equation of the evolute, we must 
find the relation of a to b in equation (1), regarding a and b as 
the co-ordinates of the centre of the circle of curvature, and 
consequently the co-ordinates of the evolute. 







EVOLUTES AND INVOLUTES. 167 

We have found (Art. 131) for the circle of curvature 

*-«=-|j-<y--&); (2) 

and combining these equations with that of the involute curve, 
we may obtain an equation from which x and y are eliminated. 

We must therefore differentiate the equation of the involute 
twice ; deduce the values of dy and d 2 y, and substitute them 
in equations (2) and (3) ; two new equations will thus be ob- 
tained involving #, b, x, and y. 

Combine these equations with the equation of the involute ; 
and eliminate x and y ; the resulting equation will contain 
only a and b and constants, and will be the equation of the 
evolute curve. 

137. To find the equation of the evolute of the common 

parabola. 

The equation of the involute is y 2 ~mx. 

_ TT . dy m _ d 2 y m 2 

Whence -r-= •$- ; and ~ri — — n- 

dx 2y y dx 2 4jr 

Substituting these values in equations (3) and (2), Art. 136, 

and reducing, we have 

y—b= — o+y: whence b 2 — — r, 

, 2y 2 m 

and x— a~— — — 77. 

m 2 

Substituting for y 2 its value mx, we have 

7o 16a? 3 _ _ m 2a— m 

o 2 =z : and x—a=— 2x—-tt: or x= — s — . 

m ' 2 ' 6 

Substituting this value of x in the preceding equation, we 
have 2 

which is the equation of the evolute, and shows it to be the 
semicubical parabola (Anal. Geom., Art. 237, Ex. 2). 



168 



DIFFERENTIAL CALCULUS. 




m 



If we make £ = 0, we have a —~k^ and 

hence the evolute meets the axis of ab- 
scissas at a distance AC from the origin 
equal to half the latus rectum. If we 
transfer the origin of co-ordinates from A 
to 0, the above equation reduces to 
16a 3 



b 2 = 



27m 



The evolute CM corresponds to the part AP of the involute, 
and CM' to the part AP'. 

138. To find the equation of the evolute of the circle. 
The equation of the involute is x 2 +y 2 =r 2 . 

/pi 



Whence 



dy x , d 2 y 
-=— ; and 



dx 



y 



dx 



2 — „,3* 



y° 



Substituting these values in equations (3) and (2), Art. 136, 
and reducing, we find a = 0, and b = 0. 

Hence the evolute is a point, viz., the centre of the circle. 
This is evidently correct, since all normals to a circumference 
meet at the centre of the circle. 



139. To find the equation of the evolute of the ellipse. 
The equation of the involute is A 2 y 2 + B 2 x 2 =A 2 T> 2 . 



Whence 



dy 



B 2 x 



, d 2 y 
and-r4=: 



B 4 



dx A 2 y' ""^ dx 2 A 2 y 3 ' 
Substituting these values in equations (3) and (2), Art. 136, 
and reducing, we find 



y- 



x—a-. 



y(Ay+BV) 
A 2 B 4 ' 
a?(A 4 y 2 + BV) 



and * — u/_ 

Now by the equation of the curve 

Ay = A 2 B 2 -BV, and B 2 # 2 = A 2 B 2 -A 2 y 2 ; 
whence Ay + PA^B^A^CV), or =A 2 (B 4 + Cy), 
putting C 2 for A 2 -B 2 . 






EVOLUTES AND INVOLUTES. 



169 



Hence, by substitution, 

whence #2 = i__i ? an( j ^2 _/___» 

Substituting these values in the equation of the involute, we 
have 

:A 2 B 2 : 



A*(*f 



A 2 B2 



If a = 6, then b=. : 




or, dividing each term by — 5-, we obtain 

(bBf + (aAf = CF = (A 2 - B 2 f , 
which is the equation of the evolute of an ellipse referred to 
its principal axes. p 2 

:-o; so that the curve meets the axis of y 

in two points P /r , P /;/ , equidistant from 

C 2 
the origin. If b = 0, then 0=±-t-; so 

that the curve also meets the axis of x 
in two points, P, P', equidistant from 
the centre. ^ 2 

If a is numerically greater than -r-, 

the ordinates become imaginary ; and if b is numerically great- 

C 2 
er than ^-, the abscissas become imaginary. Hence the curve 

is limited by the four points P, P', P", P ;// , and touches the 
axes at those points. It consists, therefore, of four branches 
symmetrically situated, as shown in the fig- 
ure, with cusps at P, P', P 7 ', and P /7/ . 

140. A normal line to any plane curve 
is a tangent to its evolute. 

Let AXP be any curve, and CTC be its 
evolute. Let T be any point in the evolute 
whose co-ordinates are a and b ; and from 
T draw a normal to the curve AP, meeting 

H 




170 DIFFERENTIAL CALCULUS. 

the curve at N, whose co-ordinates are x and y. The equation 
of this normal (Art. 90) becomes 

or x ^ a +^(y-b)=,0. (1) 

Now if the point N is moved along the curve, the corre- 
sponding point T will also be moved ; that is, a and 5, as well 
as y, are functions of x. In order to discover the change in 
each of these quantities caused by a change of x, we must dif- 
ferentiate equation (1), regarding all the quantities except dx 
as variables. We thus obtain 

d 2 y, , s dy 2 dbdy „ 

or dx 2 + dy 2 + yd 2 y— bd 2 y — dadx — dbdy = 0. (2) 

But since the point a, b, is on the evolute, we have (Art. 136) 

dx 2 + dy 2 

whence dx 2 + dy 2 +yd 2 y— bd 2 y = 0. 

Hence equation (2) becomes —dadx— dbdy —0 ; or 

db dx 
da~~ dy 
Hence equation (1), which is the equation of the normal to 
the involute at the point x, y, may be written 

a dh ( \ 

y- b =da^ x - a ^ 

which, by Art. 89, is the equation of a tangent to the e volute 
at the point &, b. Hence the radius of curvature is normal 
to the involute and tangent to the evolute. 

141. The difference between any two radii of curvature of 
a plane curve is equal to the arc of its evolute comprehended 
between them. 

Differentiating the equation 

(x-a) 2 + (y-b) 2 =r 2 , 



EVOLUTES AND INVOLUTES. 



171 



regarding all the quantities as variable, we have 

(x — a)dx + {y— b)dy — (x— a)da — (y — b)db = rdr, 
But by Art 140, (x - a)dx +{y- V)dy = 0, 

' da* + db 2 \ 2 
da 2 )- T > 



and 



Hence 



{x—aj 



a) 



and 



Jda 2 + db 2 \ rdr 



Dividing equation (2) by the square root of equation (1), we 
hav e Vda 2 + db 2 =d?\ 

But by Art. 121, \/da 2 + db 2 represents the differential of an 
arc of the evolute ; hence the differential of the radius of 
curvature of a curve is equal to the differential of the arc of 
its evolute ; or the radius of curvature and the length of the 
evolute vary by the same differences, so that the difference 
between any two radii is equal to the arc of the evolute com- 
prehended between them. 

142. Involute described mechanically. The preceding prin- 
ciples indicate how any involute may be described mechanic- 
ally from its evolute. Let ACC'C" be 
any curve, and let a thread be fastened 
to it at some point beyond Q>" . Let a 
pencil be fastened to the free end of 
the thread, and the thread drawn tight 
to the curve. Move the pencil along 
from A to P, P 7 , P 7/ , etc., keeping the 
string tight as it is gradually unwound 
or evolved from the arc ACC'C", and 
the pencil will describe a curve line 
APP'P", which is the involute of 
ACC"; for at any point as P'the por- 
tion of string unwound will coincide with the tangent at C, 
and will therefore be the radius of curvature at the point P' 




172 



DIFFERENTIAL CALCULUS. 



(Art. 140). From this mode of description the terms evolute 
and involute were derived. 



THE CYCLOID. 

143. A cycloid is the curve described by a point in the cir- 
cumference of a circle, when the circle is rolled in the same 
plane along a given straight line. 

The equation of the cycloid (Anal. Geom., Art. 243) is 
x = ver. sin. - l y — Y2ry—y 2 . (1)' 

144. Differential equation of the cycloid. The properties 
of the cycloid are most easily deduced from its differential 
equation, which is obtained by differentiating both members 
of equation (1). 

By Arts. 45 and 53, , 



d. ver. sin." 1 



y- 



By Art. 31, 



Hence 



or, reducing, 



dx 



dY2ry—y 2 
rdy 



Y2ry—y 2, 
rdy — ydy 
Y2ry—y 2 

rdy — ydy 



Y2ry- 
dx~- 



-y 2 Y2ry- 
ydy 



-W 



V2ry—y 2 

which is the differential equation of the cycloid. 



(2) 




145. Subtangentj etc., of the cy- 
cloid. If we substitute the preced- 
ing value of dx in the formulas of 
Arts. 91-94, we shall obtain the val- 
ues of the tangent and subtangent, 
normal and subnormal of the cycloid. 
If x = AR, and y = PR, we shall have 
dx y 2 



subtangent TR— y-r-= , 

& *dy y/2ry- 



y 



THE CYCLOID. 173 

tansent PT: 



j &^ 



i-W 7 



dy 



dy 2 V2ry—y^ 



subnormal 'RN=yj- ) — ^2ry — y 2 ', 

normal PN=y\/l + ^= V2^. 

The subnormal RN is equal to PH of the generating circle, 
since each is equal to V2ry—y 2 ; hence the normal PN and 
the diameter EN intersect the base of the cycloid at the same 
point. Hence, if it is required to draw a normal to the cycloid 
at a given point, draw the generating circle passing through 
the given point, and from the point where this circle touches 
the base of the cycloid draw a straight line to the given point ; 
it will be the normal required. 

146. Tangent parallel to base, etc. From equation (2), Art. 
144, we have dy V2 ry-y 2 IW~ 

dx~ y ~ V y > w 

which becomes when y=2r, and oo when y=0; hence at 
the point B, the extremity of the greatest ordinate, the tangent 
is parallel to the base ; and at the point A, where the curve 
meets the base, the tangent is perpendicular to the base. 

147. Curve concave toioard the base. Squaring both mem- 
bers of equation (1), Art. 146, we have 

dy 2 2r 
dx 2 ~~ y 
Differentiating, we have 

2dyd 2 y 2rdy d 2 y r 
dx 2 y 2 ' dx 2 ~ ~~y 2 ' 

Since the second differential coefficient is always negative, 
and y always positive, the curve is always concave toward the 
axis of x (Art, 108). 

148. Radius of curvature. Substituting the values of dy 



174 



DIFFERENTIAL CALCULUS. 



and d 2 y in the expression for the radius of curvature (Art. 



131), 
we obtain 



E: 



(dx 2 + dy 2 ) 



3 



E: 



dxd 2 y 
V V ) 



T 

But we have found, Art. 145, the normal equal to V*2ry ; 
hence the radius of curvature is equal to tioice the normal at 
the point of osculation. 

If 2/=0,E=0; and if y = 2r,~R = 4:/\ Hence the radius of 
curvature at A is equal to 0, and at 13 it is 4;\ 

149. Equation of the evolute. To obtain the equation of 
the evolute, we must substitute the values of dy and d 2 y al- 
ready found for the cycloid in equations (3) and (2) of Art. 
136. After reduction, we find 

y—b — 2y; whence y— —b. 
Also, x—a-=.—2^2ry—y 2 \ whence x = a — 2V — 2rb — o\ 

Substituting these values of x and y in the equation of the 
cycloid (Art. 143), we obtain 

a = ver.sm.~ l (-b)+V-2rb-b 2 , (1) 

which is the transcendental equation of the evolute referred 
to the primitive origin and the primitive axes. 

Produce BD to A ', 
making DA'=BD, 
and transfer the or- 
igin to A 7 , the new 
axes being A'X' and 
A'B, and the new co- 
ordinates a! and V . 
Then for any point 




m X 



as P 7 we shall have AE = a, BP' = £, 
A'M=a', MP' = 6', 
a' being negative when measured to the left of A'. 



THE CYCLOID. 



175 



' Since AD = ttv, a — wr + a' ; 

and since 

RM=-2r, RP'=-2r+&', or b=-2r+b'. 

Substituting these values in equation (1), we have 
7rr + a' = Yer.sm.-\2r-b f ) + V2rb'-// 2 . 

But nr — AD = ver. sin.-^' + ver. sin r\2r— V). 

Hence — a' = ver. sin -1 6' — V2r// — b' 2 , 

which is the equation of the evolute referred to the new axes. 
This equation is of the same form, and contains the same con- 
stants as the equation of the involute, except the sign of a'. 
But the branches A'A and A'C are symmetrical and equal, 
being derived from the equal cycloidal arcs AB and CB, and 
hence the sign of a' is indifferent; or the evolute of a cycloid 
is an equctl cycloid. 

The same conclusion might have been derived from the 
property proved in Art. 14$, that the radius of curvature is 
double the normal. Since the centre of curvature for the 
point P is on the same line as the normal, and at a distance 
equal to twice PX, it follows that XP':=XP; and hence the 
arc NP of the circle whose radius is r is equal to the arc XP' 
of a circle having the same radius, and touching AD in N. 
But the arc NP is equal to the line AX; therefore the arc LP 7 
is equal to A'L ; or the curve A'P'A may be supposed to be 
generated by the point P' as the circle LP'X rolls along LA'X'. 

If we suppose the two circles XPE,XP'L to roll along the 
bases AX, A'X' at the same rate, so as to keep their centres in 
the same vertical line, P' will describe the evolute, while P 
describes the involute. 



INTEGRAL CALCULUS. 



CHAPTER I. 

INTEGRATION OF MONOMIAL DIFFERENTIALS. OF CERTAIN BINO- 
MIAL DIFFERENTIALS. DEFINITE INTEGRALS. TRIGONOMETRICAL 
FUNCTIONS. 

150. Integral Calculus defined. In the preceding chapters 
we have shown how to differentiate a function of one or more 
variables. The integral calculus is the converse of the differ- 
ential calculus, its object being to determine the expression or 
function from which a given differential has been derived. 
Thus we have found that the differential of x n is nx n ~ l dx. 
Therefore, if we have given nx n ~ l dx, we know that it must 
have been derived from x n , or x n plus a constant term. 

The primary object of the differential calculus has been de- 
fined (Art. 17) to be to determine the ratio between the rate of 
variation of the independent variable and that of the function 
into which it enters. The object of the integral calculus, there- 
fore, is, having given the ratio of the rate of variation of a 
function to that of the independent variable ^ to determine the 
function. 

151. Integral defined. The integral of a differential func- 
tion is an expression which, being differentiated, produces the 
given differential. 

Integration is the process of deducing the integral function 
from its differential. 

152. Symbol of integration. The symbol by which the 



178 INTEGKAL CALCULUS. 

operation of integration is represented is/, a form derived from 
the old or long S. 

Leibnitz regarded the differentials of functions as indefinitely 
small differences, and he regarded the function as being the 
sum of an infinite number of these infinitesimal portions; 
hence the integral of a differential expression was denoted by 
the letter S placed before the given expression. As it fre- 
quently became necessary to place the character S before very 
complicated expressions, the symbol was elongated into the 
form/. On account of its convenience, this symbol has come 
into general use even by those who reject the philosophy of 
Leibnitz. When placed before a differential expression, it is 
to be regarded simply as denoting the expression from which 
the differential may be supposed to be derived. Thus 

fnx n ~ l dx—x n . 

153. Integration not always possible. Whatever ma3 T be 
the form of a function, its differential may always be found 
by giving to the variable an arbitrary increment, finding the 
corresponding increment of the function, and determining the 
limiting value of this ratio. There is, however, no such general 
method of integration. If we are required to integrate f(x)dx, 
we can not be sure that the process is even possible ; that is, 
we can not say that there is any function whose differential is 
t f(x)dx, unless we find that this quantity may be obtained by 
differentiating some known function. The method which we 
are therefore compelled to pursue is to collect a large number 
of integrals by examining the results of the differentiation 
of simple functions, and then by various artifices to make the 
integration of other functions depend upon integrals already 
determined. 

154. Constant factor. We have seen (Art. 19), that the 
differential of the product of a variable multiplied by a con- 
stant is equal to the constant multiplied by the differential of 
the variable. Hence we conclude that the integral of any dif- 



ELEMENTARY RULES. 179 

ferential multiplied by a constant quantity is equal to the con- 
stant multiplied by the integral of the differential. 

Thus, since the differential of Ax is Adx ? it follows that 
JAdx = Ax — Afdx. 
Hence, if the expression to be integrated has a constant fac- 
tor or divisor, this factor or divisor may be placed without 
the sign of integration without affecting the value of the in- 
tegral. Thus, fabx 2 dx = abftfdx. 

155. Sum of several differentials. We have seen (Art. 22) 
that the differential of the algebraic sum of two or more func- 
tions is the algebraic sum of the differentials of those func- 
tions ; that is, 

d(ax 3 — 2a 2 x 2 — x) = 3ax 2 dx — ^a 2 xdx — dx. 

Hence f{3ax 2 dx — 4ca 2 xdx — dx) = ax 3 — 2a 2 x 2 — x 

=/3ax 2 dx —f^a 2 xdx —fdx ; 
and generally, 

f(Adx+Bdx + ete.)=/Adx+fBdx+ etc.; 
that is, the integral of the sum or difference of any number 
of differentials is equal to the sum or difference of their re- 
spective integrals. 

156. Constant term added to the integral. We have seen 
(Art. 20) that constant terms connected with any function of a 
variable disappear in differentiation ; that is, 

d(u+C)—du; 
that is, an infinite number of expressions, differing from each 
other only in the value of the constant term or terms, when 
differentiated, will produce the same differential. Hence in 
integrating we represent any possible constant terms by an un- 
determined constant, which we usually represent by C. Thus 

that is, a constant must always be added to the integral of a 
function. 



180 INTEGKAL CALCULUS. 

157. Monomial differentials. We have found (Art. 21) that 
the differential of ax m+1 is 

a(m + T)x m dx; 
ad(x m+1 ) 



whence ax m dx = 

ax m+1 

m+1 



711 + 1 

Therefore 7 is the function whose differential is ax m dx\ 



ax m+l 

that is, fax m dx — r + C. 

' J m + 1 

Hence, to obtain the integral of a monomial differential, in- 
crease the primitive exponent of the variable by unity, mul- 
tiply by the constant factor, if there is one, and divide the 
result by the new exponent. 

Ex.1. What is the integral of 3ax 2 dx^. Ans. ax 3 + G. 

Ex. 2. The rate of variation of the independent variable x 

is to the rate of variation of a certain algebraic expression as 

ax* 1 
1 to -T-; it is required to find that expression. ax ^ 

b Ans. oT + C. 

1 3 ± 

Ex. 3. What is the integral of X s dx ? Ans. -rx^+G. 

dx — JL 

Ex. 4. What is the integral of -7-, or x z dx% 

v Ans. 2x 2 + G. 

dx 
Ex. 5. What is the integral of -3, or x~ z dx% -j 

dx 

Ex. 6. What is the integral of ax 2 dx+— — - ? 

2yx 3 

v ' ax 3 i. _, 
Ans. -^-+x* + C. 

Ex.7. What is the integral of —-x b dx% 1 

D Ans. 3^"^+C. 

Ex. 8. What is the integral of ax^dx % Sax^ 

Ans. — ^— + C. 
5 

_2 1 

Ex. 9. What is the integral of - ? ^s. + 0. 

^ a a 



n — m 



x n dx 

Ex. 10. What is the integral of — % A nx n _ 

& a Ans. —, r+C. 

a(n— m) 



ELEMEXTAKY RULES. 181 

158. Same method extended to polynomials. We have seen 
(Art. 32) that any power of a polynomial may be differentiated 
by diminishing the exponent of the power by unity, and then 
multiplying by the primitive exponent, and by the differential 
of the polynomial. Thus the differential of (ax + bx 2 ) 3 is 

3(ax + bx 2 ) 2 {adx + 2bxdx). 
Hence, in order to integrate a compound expression consisting 
of any power of a polynomial multiplied by its differential, 
increase the primitive exponent of the polynomial by unity, 
and divide the result by the new exponent. 

Ex.1. What is the integral of (a+Z<&) 2 Qxdx% 

( a + 3x 2 ) 3 ., 
Ans. v — g-^ + C. 

Ex. 2. What is the integral of (2x 3 — l)6x 2 dx t 

Ans. 
Ex. 3. What is the integral of (6— x 3 ) 3 x 2 dx ? ,„_ ^ 4 

Ans. ■ 



12 — 
Ex. 4. What is the integral of (x+ax)*(dx + adx) ? 

2 3 

Ans. ^(x+ax^+C. 

dx 
Ex. 5. What is the integral of j- ? A?is. 

Ex. 6. What is the integral of (ax 2 + bx 3 ) 2 (2ax+3bx 2 )dxl 

Ex.7. What is the integral of (ax + bx 2 )~ J (a+2bx)dx! 

dx 
Ex. 8. What is the integral of -, — — tt ? 
& {a—xf 

Ex. 9. What is the integral of -rz ^ ? 

& (1— x 2 ) 2 

x^dx 
Ex. 10. What is the integral of — 7— ? 

159. When the exponent is —1. The rule given in Art. 157 
fails when m=— l,for in this case we should have 

f* <fe =3i+i=o = o =00 ' 

which shows that the rule is inapplicable. 



182 INTEGRAL CALCULUS. 

dx 
But x~ l dx is the same as — , and we know (Art. 37) that 

x 

this expression was obtained by differentiating the logarithm 

of the denominator. Therefore 

fx~ l dx= I — =log. x+G. 
%j x 

fadx _ _. 

Also, / =a iog.x+G; 

that is, if the numerator of a fraction is the product of the 
differential of its denominator by a constant quantity, its in- 
tegral is the product of the constant by the Na/perian loga- 
rithm of the denominator. 

dx 

Ex. 1. What is the integral of — ■ — ? Ans. log. (a+x) + G. 

(&-j~X 

Ex. 2. What is the integral of r^ ? 

& a+ox* 

Ans. log. (cr+ bx 2 ) + C. 

Ex. 3. What is the integral of 7-5 ? 

Ans. — log. (0— bx 2 ) + C. 

CI Ci K* 

Ex. 4. What is the integral of -r— ? Ans. 

Ex. 5. What is the integral of — — ? 

x 

Ex. 6. What is the integral of ^ + 2x ^ dx 2 
Ex. 7. What is the integral of 



"to* 



3^ + 7* 



Ex. 8. What is the integral of sr-g — r \ J Lx% 

3x 2 dx 
Ex. 9. What is the integral of 3 , 2 ? 

g x 2 + 2# + 1 
Ex.10. What is the integral of -«-: — r^ ■r^dx'i 



BINOMIAL DIFFERENTIALS. 183 

xdx 
Ex. 11. What is the integral of , , ? 

a + bx x a 

Ans. £—filog.(a + l>x). 

x^dx 
Ex. 12. What is the integral of -r- — ; ? 

& x 2 + a 2 

• Note. — Divide the numerator by the denominator. 

x* a 2 x 2 a 4 - _ , _ 
Ans. -^ - -y + -g log. (a? 2 + a 2 ). 



160. Certain polynomial differentials. Every polynomial 
of the form {a +bx+ ex 2 + etc.) m x n dx, 

in which m is a positive whole number, may be integrated by 
raising the quantity within the parenthesis to the mth power, 
multiplying each term by x n dx, and then integrating each 
term separately. 

Ex.1. What is the integral of (a + bx) 2 dxt -, 2 3 

Ans. a 2 x + abx 2 +—Q--\-C 

Ex. 2. What is the integral of (5 + 7^ 2 ) 2 ^ ? 

Ans. 25x + — 5- + -— + C. 

Ex. 3. What is the integral of (a+3^ 2 ) 3 ^? 
Ex. 4. What is the integral of {a-\-bx?) 2 x 3 dx% 

Ex. 5. What is the integral of {a—x 2 ) 3 x 2 dx ( i 

i _i 
Ex. 6. What is the integral of {a—bx^fx ' 6 dx\ 

Ex. 7. What is the integral of {a + bx 3 ) 2 x 2 dx ? 

Ex. 8. What is the integral of (a-\-bx 3 fx*dx% 

Ex. 9. What is the integral of (a 2 + bx*) 3 x 5 dx < i 

161. Particular binomial differentials. By the introduc- 
tion of an auxiliary variable, many binomial expressions may 
be transformed into monomials, and then integrated as in 
Art. 157. 

Let du = {a + bx 1l ) m x n ~ l dx, 

in which the exponent of the variable without the parenthesis 
is one less than the exponent of the variable within. 



184 INTEGRAL CALCULUS. 



Let us put 


y—a + bx n . 


Then 


dy=bnx n ~ 1 dx, 


and 


dy 

X CLX'^-'j • 

on 



y m dy 
Hence ~bn ' 

and integrating, we have 

u= , , ,w +C; 
(m + l)on 

and replacing the value of y, we have 

(a + bx n ) m + l 



u- 



+ C; 



(m+l)bn 

that is, to integrate a binomial differential when the exponent 
of the variable without the parenthesis is one less than that 
within, increase the primitive exponent of the binomial by 
unity, and divide the result by the product of the new expo- 
nent,the coefficient, and the exponent of the variable within 
the parenthesis. 

Ex.1. What is the integral of {a + Zx^fxclx^ 

(«+3sy Ja+to*r 
J±n8 ' 4.3.2 + 24 + 

1 

Ex. 2. What is the integral of {a + bx^mxclxl 

7)1, 3 

Am. ^(a + bx^+C. 

xdx 
Ex. 3. What is the integral of . ? 

Va +x Am. Vtf+tf+C. 

3 

Ex. 4. What is the integral of (a + bx 2 ) 2 exdx ? 5 

. e(a+bx 2 f _ 
Am. 5b + 0. 

Ex. 5. What is the integral of (tf+x'fanx^dxt. 

_i 
Ex. 6. What is the integral of (3 - 4« 5 ) 2 2x*dx ? , 

(3-4*2* 
Am. — - =— i -+C. 



BINOMIAL DIFFERENTIALS. 185 

162. Binomial differentials. Every binomial differential 
of the form du = A(a + hx) m x n dx 

can be integrated when either ?n or n is a positive whole num- 
ber, and the exponent of x in the parenthesis is unity. 

If m be a whole number and positive, we may integrate as 
in Art, 158. 

If n be a whole number and positive, m being either frac- 
tional or negative, place 

y=a+bx; 

. y—a 

then x=*-t— , 



<fc=-|, du^A^-^jy^, 



which may be integrated as in Art. 158. If we replace the 

value of y, the integral will be expressed in terms of x. 

1 
Ex.1. What is the integral of x 2 (a + xy I dx r i 

Put y=a + x; then x=y—a, dx — dy^ 

1 2 7 4 5 2 3 

2 3 35 

or, replacing the value of ?/, 

2 3 8 

w = ^M + x) 2 (5a? 2 - ±ax + »a 2 ) + C. 

xdx 
Ex. 2. What is the integral of — r ? 

Place the expression under the form 

du=x(a + bx) ' 2 dx; whence du— ^ y ^dy. 

?y"^/2 \ 2 i 

Hence ^=*^(-y— 2^)=-qr^(^+te) 2 (^— 2a) + C. 



186 



INTEGRAL CALCULUS. 






Ex. 3. What is the integral of 



x 2 dx 



(a+ix) 
Ans. jiia+bxY 



~(a+bx) + a 2 \ + C. 



Ex. 4. What is the integral of 



ils (JLJU 



Am. ^(l+x)^—^(l+xf+2x{l+xf+G. 



Ex. 5. What is the integral of 



tL/iJuJU 



(a + hx) 



Ans, 



3. T s/a+lx a\ „ 



163. Geometrical illustration. The problem of finding the 
areas of curves was one of those which gave rise to the integ- 
ral calculus, and affords an illustration of the preceding prin- 
ciples. Let OMP be a parabola whose 
equation is y 2 — mx^ and suppose it is re- 
quired to find the area included between 
this curve and the axis of X. Let A 
represent the required area. Then (Art 




or 



122) dA=ydx. But for the paraboia dx. 
dK-ydx^ 



%ydy 
m 



; hence 



m 



2y 



Therefore A = ^- + C = ^-(mx) * + C = ^rn^x 1 + C, 



6m 



3m} 



a) 



which is a general expression for the required area. 



164. Definite integral. To complete each integral as deter- 
mined by the preceding rules, we have added a constant quan- 
tity C. While the value of this constant is unknown, the ex- 
pression is called an indefinite integral ; but in the application 
of the calculus to the solution of particular problems, the value 



DEFINITE INTEGRAL. 187 

of the constant C may be determined from the conditions of 
the problem. When there is any thing in the nature of the 
problem from which we can know the value of the function 
for some particular value of the variable, by substituting this 
value in the indefinite integral the value of the constant C can 
be found ; and since C is constant, if we find its value for a 
particular value of the variable, we know its value for all 
values of the variable. 

When the value of the constant has been determined, and a 
particular value assigned to the independent variable, the value 
of the integral is known, and it is called a definite integral. 

To determine the value of C in the example of Art. 163, we 
perceive that if we estimate the area from O, the integral 
which expresses the required area becomes when x = 0. 
Hence equation (1) becomes 

= + 0. 
Therefore the constant C is equal to 0, and the entire integral 
is 2v 3 2 x 3 

3m 3 

This represents the area estimated from O to any value we 
may choose to give to x. If x=QH, A = area OPR. 

If, however, we estimate the area from the ordinate DF, 

77b 

which passes through the focus F, calling A = when x— -r-, 
we have 



m 2 



_ 2 x(m\* _, m 2 „ 



Hence C = — t-^-, and the corrected integral is 

. 2 i| m 2 

A =r i2x -i2- 

This represents the area estimated from the focal ordinate DF 
to any value we may choose to give to x. If # = 011, this cor- 
rected integral represents the area FDPR. 

165. Integral between limits. Suppose we wish to obtain 
an expression for the curvilinear area MNRP, contained be- 



188 INTEGRAL CALCULUS. 

tween the two ordinates MN, PR. We first obtain the area of 
the portion OMN. Suppose the variable x to be equal to ON, 
which we will represent by a; then the area of OMN will be 
expressed by o x 3 

Next suppose the variable x to become equal to OK, which we 
will represent by b; then the area of OPR will be expressed by 

O i 3 

-m^b^+C. 

Subtracting the former expression from the latter, we obtain 
the value of the curvilinear area MNRP, 

i 
2 l A 2 i J 2;?& 3 " 3 .3 

Hence we find that the constant C may be made to disappear 
by giving two successive values to the independent variable, 
and taking the difference between the two integrals corre- 
sponding to these values. 

When we take the excess of the value of an integral when 
the independent variable has become equal to £, above the 
value when it was only equal to a, we are said to integrate be- 
tween the limits of x = a and x=b. 

The integration between the limits a and b is indicated by 

, . Cb 

the sign 



■s j 



Ja 



Cb i 1 2m 2 3 3 
Thus J ?n^x 1F dx=:-^-(b^-a 1F ). 

Ex. 1. Integrate / 2xdx, and illustrate the case by a geo- 
metrical example. Ans. 

Ex. 2. Integrate / 3x 2 dx ; illustrate the case by a geomet- 
rical example, and determine the numerical value when a = 4 
and 5 = 6. Ans. 



TRIGONOMETRICAL FUNCTIONS. 189 

n 
Ex. 3. Integrate / ^xdx ; illustrate the case by a geomet- 
rical example, and determine the value of the definite integral 
when a — 2 and ft = 3. Ans. 

Ex. 4. Integrate / ^x 2 dx ; illustrate the case by a geomet- 
rical example, and determine the value of the definite integral 
when a — 4 and ft = 6. Ans. 

Ex. 5. What is the value of / 2(e + x)dx, when a — 10, ft = 20, 

and £=4? and illustrate the example by a geometrical figure. 

Ans. 

3(e+?ix 2 ) 2 2nxdx^]ien a — ^ 
6 = 6, 6 = 4, and n — 2% Ans. 

Jrh dx 
— — .when a = 2, # = 3, and 
a e + ar 

6 = 4? Ans. 



INTEGRATION OF THE DIFFERENTIALS OF TRIGONOMETRICAL 

FUNCTIONS. 

166. Integration of elementary trigonometrical expressions. 

(1) We have found, in Art. 44, that if x represents an arc of 
a circle, d sin. # = cos. xdx. 

Hence /cos. xdx = sin. x + C. 

(2) We have found, in Art. 45, that d cos. x— —sin. ajrfa?. 
Hence /sin. a%?# = — cos. x + 0. 

(3) We have found, in Art. 45, that d ver. sin. #=sin. xdx. 
Hence /sin. #*?# = ver. sin. x + C. 

(4) We have found, in Art. 46, that d tang. #=sec. 2 t£#fe, or 

cos.V r , 

Hence fsec. 2 xdx, or / =tang. x + C 
y ' J cos. 2 # 5 



190 INTEGRAL CALCULUS. 

(5) We have found, in Art. 47, that d cotang. x= — cosec. 2 xdx, 
dx 

Or : =—. 

sin. 2 x 



C dx 
Hence / - — —= — cotang. x+C 

%J SID. X 



(6) We have found, in Art. 48, that d sec. x = tang, x sec. xdx. 
Hence /tang, x sec. xdx = sec. a? + C. 

(7) We have found, in Art. 49, that d cosec. x= —cotang. x 
cosec. xdx. 

Hence /cotang. x cosec. xdx = — cosec. # -f C. 

167. Integration of more complicated expressions. Ex- 
pressions which do not appear to be of the form for the appli- 
cation of the previous methods may sometimes be brought to 
the required form by reduction. 

Ex. 1. Find the integral of cos. nxdx. 

/cos. nxdx — - /cos. nxdinx) = - sin. nx + 0. 

Ex. 2. Find the integral of sin. {m + n)xdx. 

cos. (m + n)x _, 

Ans. v , ; +C. 

m+n 

Ex. 3. Find the integral of tang. xdx. 

r . 7 Cmv.xdx r—dcos.x -, . ^ 

/tans. xdx = I = / =— log.cos.#+C. 

J COS. X J COS. X 

Ex. 4. Find the integral of cotang. xdx. 

Ans. log. sin.#+C. 

dx 
Ex. 5. Find the integral of =-; . 

° 1 + COS. X 

By Trig., Art. 74, 

1 + cos. * = 2(cos. Ixf. 
Hence 

f dx. r dx _ r dgx) 

J 1 + cos.x J 2{cob.^x) 2 J (cos. \xf b '^ T ' 

by Art. 166, (4). 



Ex. 6. Find the integral of 



TRIGONOMETRICAL FUNCTIONS. 191 

sin. xdx 



° 1 + COS. X 

/s\\\.xdx C— <#(l + cos. X) 1 M . \ t r\ 

—= I — -^ — — —!-= — log. (l + cos.#) + C. 
1 + COS.CC J 1 + COS. X 



dx 
Ex. 7. Find the integral of 



sm. a? cos. x 



f * = r sec - 2 ^= r rftan g^=log.taag. 8 +C. 

*/ sin. x cos. a? t/ tang, x J tang. a? 



6?a? 
Ex. 8. Find the integral of - 

& ft! 



sin. a? 

By Trig, Art. U 2 ^ 

sm. x — 2i sm. \x cos. -kB=-7 , X9 . 

J J (sec. \xf> 

dx (sec. \x) 2 dx (sec. -|aj) 2 <^(-^c) <i tang. ^ 

sin. x~ 2 tang. -Ja? ~ tang. \x ' tang. -^a?" 

Therefore / -; = log. tang. ia?+C. 

J sm. a? 

Ex. 9. Find the integral of tang. 2 xdx. 

/tang. 2 xdx=f(sec. 2 x— l)dx— tang, a?— a?+C. 

Ex. 10. Find the integral of sin. 2 xdx. 

By Trig., Art. 74, sin. 2 a? = 5 — ~ cos. 2a?. 

Hence sin. 2 xdx = ——- cos. 2a?<fe. 

1 a? 1 

Therefore /sin. 2 a%fe:=ia?— t/cos. 2xd(2x)=-^— j sin. 2x+G. 

168. Differential of arc in terms of sine. We have found 
(Art. 53) that if ?/ designates an arc, and x its sine, the radius 
of the circle being unity, 

dx 

dy=- 



Vl—x 2 

Hence / — y+C^sin'^+C. 

J Vl—x 2 



192 INTEGRAL CALCULUS. 

If the arc be estimated from the beginning of the first quad- 
rant, the sine will be when the arc is 0, and consequently 
equals zero. 

Therefore the complete integral is 



/dx 
VTZc 



:the arc of which x is the sine. 



VT— x* 
If it be required to integrate an expression of the form 

Va 2 —x 2 
it may be done by the aid of an auxiliary variable. 

Assume v = -, or x — av ; 

then dx = adv, and Va 2 —x 2 = a Vl—v 2 . 

Substituting these values in equation (1), we have 

dv 



'3 



Vl — v 
Hence y=the arc whose sine is -y=the arc whose sine is - ; 



that is, f dx . x 

. r_ SlU." 1 - 

Ya 2 —x 2 a 



/dx 
Va 2 — x 2 & 

dx 



Ex.1. Integrate the expression dy— 



Ex. 2. Integrate the expression 



V4— x 2 

'C 

5dx 



Ans. y~ arc whose sine is ^ + C. 



V3-x 2 

Ans. 5 sin. -1 — 7^ + C. 



Ex. 3. Integrate the expression , 

5 ^ V3-2a? 2 



3 /2 

J.^5. —777 sinr'ajyo + C. 



V2 OXA " "V 3 
169. Differential of arc in terms of cosine. We have found, 



TEIGONOMETEICAL FUNCTIONS. 



193 



(1) 



in Art. 53, that if y designates an arc, and x its cosine, the ra- 
dius of the circle being unity, 

— dx 

Hence / =y+C = cos.- 1 a?+C. 

To determine the constant C, we observe that if the arc be 
estimated from B, the beginning of the first c 
quadrant, the cosine becomes when the arc 
becomes a quadrant, which is represented by 

jr. Making, therefore, / — == = - and x = 

^ J Vl—x 2 * 

in equation (1), we have A 

7T 

- = arc whose cosine is + C; (2) 

7T 

and since the arc whose cosine is is equal to ^, Eq. 2 becomes 

2-2 + °- 
Hence C = 0, and the complete integral is 

— dx 




/; 



Vl—x 2 

If it be required to integrate an expression of the form 

—dx 



Va 2 - 



df 



it may be done by the aid of an auxiliary variable, as in Art. 
168, and we shall find 

— dx x 



h 



Va 2 -x 2 
Ex.1. Integrate the expression 



: cos." 



a 

-2dx 



V±z 



7f 



X 



Ans. 2 cos.-^-hC. 



Ex. 2. Integrate the expression 



— dx 

V»-i?" 1 2x 

Ans. 5 cos. -1 -o- + C. 



194 INTEGRAL CALCULUS. 

170. Differential of are in terms of versed sine. We have 

found, Art. 53, that if y represents an arc, and x its versed sine, 

dx 
dy=- 



V%x—x 2 

/dx 
. = ver. sin." l x + 0. - 

Y 2x— x 2 

If the arc be estimated from the beginning of the first quad- 
rant, we shall have C=0, and the complete integral is 

:the arc of which x is the versed sine. 



/ 



Vvx—xi 2 

If it be required to integrate an expression of the form 

V 2ax— x 2 

it may be done by the aid of an auxiliary variable. 

x 
Assume v = -, or x — av : 

then dx—adv. 

Substituting in equation (1), we have 

, adv dv 

V ~ V2a 2 v-a 2 v 2 ~ V2v-v 2 

Hence v=the arc whose versed sine is v. or - ; that is, 



/ 



dx . x 

r — ver. sin. -1 -. 

V2ax— x 2 a 

5dx 



Ex. 1. Integrate the expression 

* ^ V4#-2;s 2 g 

^4n«9. — 7^ ver. sin. -1 #. 

Ex. 2. Integrate the expression r. 

r V6^-9^ 2 x 

-4/i«§. o ver. sin.~ l 3x. 

Ex. 3. Integrate the expression =. 

V9#-4^ 2 3 8 ^ 

J.^5. o ver. sin. -1 -^-. 



TRIGONOMETRICAL FUNCTIONS. 195 

171. Differential of arc in terms of tangent. We have 
found, in Art 53, that if y designates an arc, and x its tangent, 

_ dx 

Hence J jqp^tang.-'a + C. 

If the arc be estimated from the beginning of the first quad- 

/dx 
1 ,2 — 0? hence C = and 

the complete integral is 

/ 2 = the arc of which x is the tangent. 

If it be required to integrate an expression of the form 

d y=-&+& (1) 

it may be done by the aid of an auxiliary variable. 

Assume v = -, or x = av ; 

then dx — adv. 

Substituting in equation (1), we have 

adv 1 dv 



~a 2 +a 2 v 2 a 1+v 2 ' 

1 x 

Hence y = - x arc whose tangent is v, or - ; that is, 



/dx 1 _^ 

a 2 +x 2 ~~a =>" a" 



Ex.1. Integrate the expression 



2+ 3a? 3 ' 



f dx 1 f dx 1 1 _,_#_ 1 , 

^ 2 + 3« 2 -37 f + ^ 2 ~3 X Vf tang " VI~V6 tang - *™ 

Ex. 2. Integrate the expression A , ^ „ . 1 ,3a? 

^ 4 + 9z 2 ,4^. - tang - ^ 



196 INTEGRAL CALCULUS. 

//Of* 

Ex. 3. Integrate the expression 1 - 2 . 

^.^5. -yr- tang r l x<\/5. 

172. Differential of arc in terms of secant We have found, 
Art. 53, that if y represents an arc, and x its secant, 

y dx 

d y= r^—T ' 

xvx 2 —l 

Hence / — . —sec.'^ + C. 

J xvx 2 —l 

If the arc be estimated from the beginning of the first quad- 
rant, C = 0, and the complete integral is 



r dx 

J xVx 2 —l 



If it were required to integrate an expression of the form 

., dx 

dy-- 



xVx 2 —a 2 ' 



we may proceed as in Art. 168, and we shall find 

dx 1 x 



h 



xVx 2 —a 2 a a 

Ex. 1. Integrate the expression — , 

/2 
V3°" "V 3' 

Ex. 2. Integrate the expression 



1 i / 2 

Ans. —7K sec. #y « 



x 



VZx 2 -± 



a i V3 

Ans. see." ^~k~- 



RATIONAL FRACTIONS. 197 



CHAPTEK II. 

RATIONAL FRACTIONS. INTEGRATION OF CERTAIN IRRATIONAL 
DIFFERENTIALS. 

173. By the term rational fraction is to be understood a 
fraction in which the exponents of the variable are all positive 
integers. If the numerator contain powers of x as high or 
higher than any in the denominator, we may, by division, re- 
duce the fraction to the form of an entire and rational part, 
plus a fraction in which the numerator is of a lower degree 
than the denominator. The former can be integrated by 
methods already explained, and therefore we need only con- 
sider those rational fractions in which the highest exponent of 
the variable in the numerator is at least one less than the 
highest exponent in the denominator. In order to integrate 
such a fraction, we resolve it into the sum of a number of 
fractions with simpler denominators, called partial fractions. 
This can always be done by the method of undetermined co- 
efficients explained in Algebra, Art. 375. There may be four 
cases : 

I. When the denominator of the given fraction can be re- 
solved into n real and unequal factors of the first degree. 

II. When the denominator can be resolved into n real fac- 
tors of the first degree, and some of them are equal. 

III. When some of the binomial factors are imaginary, but 
no two are equal. 

IY. When some of the binomial factors are imaginary and 
equal. 

174. Case I. When the denominator can be resolved into 
n real and unequal factors of the first degree. 

Ex.1. Integrate the expression -g — g- 5 or —, ^r. 



198 INTEGRAL CALCULUS. 



The fraction —, ^ can be decomposed into *r \ z—z f • 

x(x— 2) r 2 I #—2 a? ) 

__ r dx 1 /• j dx dx) 

Hence J ^2) = 2J \ ^2~¥ f 

a?— 2 
=i{log. (»-2)-log. a;} =| log. — -. 

Ct (ilP If* — — ^7 

Ex. 2. Integrate the expression -§ . Ans. log. . 

#2 — CIX X 

Ex. 3. Integrate the expression x ,^ +8 , or (a , + ^ + 2y 

a; A B 

Assume ( aJ +4)( flJ +2)=5+i + 5+3- (1) 

Clearing of fractions, we have 

x=Ax+2A+Bx+4B. 
Since this equation is true for all values of x, by the principle 
of undetermined coefficients we may put the coefficients of the 
like powers of x in the two members equal to each other, and 
we have 1=A + B, and = 2A + 4B; 

whence A = + 2, and B= — 1. 

Substituting these values in equation (1), we have 
x 2 1 



Hence 



(x+4)(x + 2) X + 4: x + 2' 
xdx C 2dx C dx 



/xdx _ r2dx r 
(X + 4:)(X + 2) = J x + I" J i 



(x + ±)(x + 2) J x + 4, J x + 2 

= 2 log. (a + 4)-log. (* + 2)=log. { ~^-. 

If there are n factors in the denominator of the given ex- 
pression, there will be n corresponding fractions of the form 

— ; — , — -7, etc. ; and when these are reduced to a common 

x + as x + v 

denominator with the first member, they will give an identical 
equation in which the coefficients of the like powers of x in 
the two members are equal, and hence the values of A, B, etc., 
may be determined. Hence, to decompose the fraction into 
partial fractions and integrate it, we have the following rule : 
Assume the primitive fraction (omitting the factor dx) 



RATIONAL FRACTIONS. 199 

equal to the sum of as many partial fractions of the form 
— — as there are factors of the first degree in its denom- 
inator ', the denominators being the several factors of the given 
denominator. Clear the equation of denominators ; equate 
the coefficients of the like powers of x, and thence determine 
the values of 'A, B, etc. Substitute these values of A, B, etc., 
in the assumed fractions, multiply by dx, and take the sum 
of the integrals of the partial fractions. 

Ex. 4. Integrate the expression -j — ~ fl . 

. x—3 
Ans. a log. ~ + C. 

X — jU 

Jix. 5. integrate the expression k — . 

& r x— x 6 

The factors of the denominator are 1 + a?, 1— a?, and a?. 

l + 3z 2 ABC 

Hence we assume t = t~ 1 — +^j + — . 

x— x 3 1+x 1—x x 

Clearing of denominators, we have 

1 + 3£ 2 = A#-Az 2 + B#+Bz 2 + C-C£ 2 ; (1) 

whence 3 = -A+B-C; 0=A+B; 1=0. 

Therefore A = - 2, and B = 2. 

_ r(l + 3x 2 )dx rt r dx ^ C dx rdx 

Hence I ^^- — ^—=-2 r — + 2/ . + / — 

J x—x 3 J 1 + x J 1—x J X 

= —2 log. (1 + a?) — 2 log. (1— x) + \og. x 

X 
= -log. (l-£ 2 ) 2 + log. X = \0g. ( 1 _dy +0- 

In some cases it will facilitate the finding of the values of 
A, B, C, etc., if we merely indicate the factors in clearing the 
equation of fractions ; and then, since equation (1) holds true 
for every value of x, give to x such values as to render the 
factors in turn equal to zero. 

Thus in the equation above we have 

l + 3^ 2 =A^(l-^) + B^(l + ^) + C(l+^)(l-^). 



200 INTEGRAL CALCULUS. 

Making x = 0, we have 1 = C ; 



" x=l, 


a 


4=2B; 


" a?=~l, 


u 


4=-2A; 


the same values as above. 







_ ^ _ _ . (a 3 + bx 2 )dx 
hx. 6. Integrate the expression — , =— -. 



J_^s. a log. a?— — ^— log. (a 2 — x 2 )+G. 



_ . .._ _ . (3x—5)dx 

Jix. 7. Integrate the expression 2 _^ i q - 



^^-^ i",-;: «_;'■<»-*)■ 



^L^6'. o log. (#—4) — -^ log. (a?— 2) = log 



^ ^ (a-2)* 

Ex. 8. Integrate the expression — —j-. Ans. - log. — rr- 

& r # + te a fe #+te 

T» A T j. L ll . (2# + 3)d# 

±Lx. y. Integrate the expression 3 2 o • 

5 13 

J./W. g log. (aj— 1)— - log. (a; + 2)- o log. x. 

175. Case II. When some of the binomial factors of the 
denominator are real and equal. If all the factors of the 
denominator are equal, the denominator will be of the form 
(#—■#)*, and the fraction may be integrated as in Art. 158. It 
is only necessary, then, to consider the case in which a portion 
of the factors are equal. It will be found by trial that the 
method of the preceding article is not applicable in this case. 

Thus, take the example 7 ^7 — -f v in which two of the 

[X jUJ [X — ±) 

factors are equal to x—2. According to the method of Art. 
174, we should have 

x 2 +x A B C A+B C 

(x-2) 2 (x-l)~x-2 + x-2 + x-l'- x-2 + oJ— 1? 

where A + B must be regarded as a single constant. Reducing 

the two members to a common denominator, and placing the 

coefficients of the like powers of x in the two members equal 



RATIONAL FRACTIONS. 201 

to each other, we shall have three independent equations con- 
taining but two unknown quantities. 

We may obviate this difficulty by assuming the given frac- 
tion equal to ABC 

(x-2) 2 + x-2 + x^l' 
Clearing of denominators, we have 

x 2 + x = Ax-A+Bx 2 -3Bx+2B + Cx 2 -±Cx + 4:C; 
whence 1=B + C, 1 = A-3B-4C, = -A + 2B + 4C. 
Therefore A = 6, B = - 1, and C = 2, 

x 2 + x 6 1 2 

a (x-2) 2 (x-l)~ (x-2f"^2 + ^l' 

tt f (x 2 + x)dx r 6dx r dx r 2dx 

ence J (x-2)Xx-l)~J (x-2) 2 Jx^2 + J^l 

= -^2~log- (a-2) + 2 log. (m-1) 

6 (x-1) 2 

~ a>-2 + og " x-2 ' 
If the denominator of the given fraction contain n equal 
factors of the first degree, we must assume n partial fractions 
of the form 

A B C_ _N_ 

(x-a) n + (x-a) n - 1 + (x--a) n - 2+ * ' ' + x-a 
Having determined A, B, C, etc., each partial fraction is to be 
multiplied by dx, and may be integrated as in Art. 174. 

Ex. 2. Integrate the expression -, — -7^=7 — r^r. 
& r (x + 2) 2 (x+l) 

A 2 .1 * +2 

AnS '--^V2 + l0 &^VY 

dx 
Ex. 3. Integrate the expression / j-ftw — IV 

1 _ . x—1 

Arts. ^7 — r^+-g-l°2:- — r~^- 
3(#+2) d 6 x+2 

^ , T , . (x 2 —2)dx (x 2 -2)dx 

Ex. 4. Integrate the expression x ±_^ , or X ^ X _ X) • 

12 



202 INTEGRAL CALCULUS. 

x 2 -2 A B C D 
Assume -57 T\ : =~3 + ~i + ~ + 



x 3 (x—l) x 3 x 2 x x— V 
whence # 2 -2 = A#— A-f-B£ 2 -B#+Cte 3 — Cx 2 +J)x 3 ; 
= C+D, 1 = B-C, 0=A-B, -2=-A; 
A=2, B=2,0=1,D'=-1. 
x 2 -2 2 2 11 



Hence 



x 3 (x— 1) a? 3 a? 2 # x—l J 

/(x 2 —2)dx_ C2dx C2dx Cdx __ /* cfe 
# 3 (#— 1) */ # 3 J OJ 2 J X J x—1 



1 2 

-+log.a?--log.(a>-l) 



x 4 x 



\ + 2x x 



a? 2 & * #- 

Ex. 5. Integrate the expression — 302 • 

10 (x-3\™ 

Ans. -+log.^j . 



Ex. 6. Integrate the expression , , , 3 



a+2bx 
Ans. 



~2b\a+bx) 2 ' 

176. Case III. TTA^w. 5^m6 of the binomial factors ofths 
denominator are imaginary but unequal. It is shown in 
Algebra, Art. 443, that imaginary roots are found only in 
pairs; hence the product of each pair of imaginary factors 
will be of the form (x— a) 2 -\-b 2 . Instead of assuming a partial 
fraction for each imaginary factor, we assume for each pair of 
imaginary factors a partial fraction of the form 

A#+B 
(x—a) 2 + b 2 ' 
By clearing of denominators, and proceeding as in the last 
Article, the values of A and B may be determined. 



RATIONAL FRACTIONS. 203 

_, . _ . . x 2 dx x 2 dx 

Ex.l. Integrate the expression -o- — =-; r-r,or , „ „ 1W — — rr, 

8 v x 3 + x 2 + x+V (x z + l)(x + iy 

where x 2 + l is a quadratic factor whose roots are imaginary. 
a? 2 A#+B C 

Assume (^+ix^i) = ^+r+^TT- 

Clearing of fractions, we have 

# 2 =A£ 2 + A#+B#+B + Q£ 2 + C; 
whence 1=A+C, 0=A + B, 0=B+C; 

A__ i tc_ i n_ i . 

£ 2 # — 1 1 

(^ 2 +l)(^ + l) :=: 2(^ 2 + l) + 2(^-|-l) ; 

/# 2 <fe _ 1 r xdx j r dx . r dx 

=4 log. (z 2 + l)— ^ tang.-^+i log- (a + 1). 



Ex. 2. Integrate the expression 



# 3 — # 2 +#— 1" 



A . . V^ 2 + l . ' . 

J.ng. — ^log ge _ 1 — ^tang. 'a?. 



_ ^ _ _ . (x 2 —x+V)dx 

Ex. 3. Integrate the expression a?3+a? 2 +a?+r 

J./Z.S. log. ~i—h tang." #. 

2 + l)* 

Ex. 4. Integrate the expression . 2 + 1 x^ 2+4 y 

J_fts. ^tang. -1 #— -^-tang. -1 ^. 

177. Case IV. When some of the binomial factors of the 
denominator are imaginary and equal. In this case the prod- 
uct of each pair of imaginary factors is of the form (x— a) 2 + b 2 , 
and if there are n such factors, the denominator will contain a 
factor of the form {(x— a) 2 + b 2 ) n . In order to determine the 
partial fractions, we combine the methods employed in the 



204 INTEGRAL CALCULUS. 

second and third cases, and assume n partial fractions of the 

form A#+B Cx + T> Mx+N 

{(x-a) 2 + b 2 \ n ' {(x-af + b*}"- 1 ' ' ' ' (x-a) 2 + b 2 ' 

By clearing the equation of denominators, and proceeding as 

in Art. 176, the values of A, B, C, etc., may be determined. 

_ . (x 3 +x— l)dx 

Ex. 1. Integrate the expression — , 2 ■ o\2 — • 

x 3 + x-l A# + B Cx+T> 
Assume (rt»+-2)» "(«".+ 2)» + x 2 + 2' W 

Clearing of fractions, we find 

x 3 + x-l = Ax+B + Gx 3 + J)x 2 + 2Cx + 2T>; 
whence 1=C, 0=D, 1 = A + 2C, _1 = B + 2D. 
Therefore A=-l, B=-l, G=l, D = 0. 

Substituting these values in equation (1), and multiplying 
by dx, we have 

/(x 3 +x—l)dx r(x + l)dx f xdx 

(x 2 + 2f =~J (x 2 + 2) 2+ J tf+2 

/xdx C dx r xdx 

(x 2 + 2) 2 ~~J (x 2 + 2) 2 + J ¥+2 

1 r dx t _ , * N 

= 2(^+2)-^ pT^ + * l0g - ( * +2) - 
The term , 2 1 o\2 can b e integrated by means of formula D, 

given in Art. 195. 

x dx x d t 

Ex. 2. Integrate the expression j^_ x ^ + 1 | 2 ' or ( a j_ 2a . +2 )r 

Assume ( tr 2_ 2a . + 2)2-(a ! 2_2* + 2) 2 + a; 2 — 2«+2 - (1 ) 
Clearing of fractions, and proceeding as in the last example, 
we find A = 2, B= -4, 0=1, and D = 2. 

Substituting these values in equation (1), and multiplying 
by dx, we have 

x 3 dx _ 2xdx 4dx xdx 2dx 

(x 2 -2x+2) 2 ~(x 2 -2x+2)~(x 2 -2x+2) 2 x 2 -2x+2 x 2 -2x+2' 



IRRATIONAL DIFFERENTIALS. 205 

For convenience, substitute z for #— 1, and we have 

xhlx . 2(z+l)dz 4:dz {z + l)dz 2dz 



(x 2 -2x + 2) 2 ~ (z 2 + iy (z 2 + l) 2 ^ z 2 + l ^s 2 +l 
2zdz 2dz zdz 3dz 



~~0 2 + l) 2 (z 2 + l) 2 + z 2 + l^z 2 + V 
Hence 

/# 3 <^# i ^ r dz , , _ ,„..«.,« 
p=3^w = ~^+i"y w +i iog - (s +i)+3 tang - *• 

The integral of the second term is found by formula D, 
Art. 195. 

From the preceding discussion we infer that all differentials 
which are rational fractions can be integrated, provided we 
can discover the factors of the denominator. These integrals 
can all be found by methods already explained, with only one 
exception, to be considered in Art. 195. 

INTEGRATION OF CERTAIN IRRATIONAL DIFFERENTIALS. 

178. When the irrational jparts are monomials. In the 
preceding articles, methods have been explained by which 
every rational differential can be integrated, with the excep- 
tion of one case, referred to in Art. 177. We hence infer that 
we can also integrate every irrational differential which can 
he made rational in terms of a new variable. 

The simplest irrational differentials are those in which all 
the irrational parts are monomials. Such expressions may be 
rendered rational by reducing the fractional exponents of the 

variable a? to a common denominator n, and substituting z n for x. 

i 

Ex.1. Integrate the expression ^dx. 

1—x^ 
The least common denominator of the exponents of x is 6. 
Hence we must substitute z 6 for x. Therefore dx=6z 5 dz. 
By substitution in the given expression, we have 

l-*>* 1-z 3 n fi7 6z 8 -6z 5 7 

T dx — . 5 . K)Z b dz — — 5 — T~dz. 

-, i 1 — z 2 z 2 — l ' 

1—x 3 



206 INTEGRAL CALCULUS. 

which, by division, becomes 

Hence 

r/l-x\ _ „ ( z> s 5 z* z 3 z* , , „ 1 
^W) 1 7 + 5 "4 +3 -2 + ^ l0 g- ( 1+S ) } 

7 A 2 JL 1 
„ ( X* , a? 6 cc 3 " , a? 2 SC 3 " , 1 _ ^ , 1 ) „ 

=6 i y+ 5 "t+3 - 2 +aj lo «-( 1+a? > J +a 

^ i 

Ex. 2. Integrate the expression — . dx. 

1+as 3 " 

Assume x=z 6 ; then, by substitution, we have 

x*—2x* 6(s 8 -2s 1 )dz 

i -dx— .. 2 , 

which, by division, becomes 

6(z 6 -2zs-z* + 2z 3 +z*-2z-l + ^±^\dz. 

Hence 

rx*-2x* , rt fs 7 2^ 6 s 5 2s 4 s 3 2s 2 . , . ^ ,1 

J ■ r • <&=6(^-g~g4- ^ +3 - 2"- s+log. (s 2 +l)+tang.- 1 s| 



1+x* 

7 5 2 1 



T/wO nfl /j»6 /vjO /V>2^ 11 1 A J 

: 6 1 Y ~3~ 5" + 2" + 3"" * 7 -^+log. (aj'+lJ+tang.-V j- 



1 

Sx^dx 
Ex. 3. Integrate the expression — r 



2* 

2# T — X s 



5 2 1 



^. + ^. + ^- + 4a^ + 16aj T + 32 log.(2-^)j + C. 

1 2 

Ex. 4. Integrate the expression 1 — . dx. 

5x~* 

3 * 2 4 ~ 

Am. tk x — * x + U. 
10 5 



IRRATIONAL DIFFERENTIALS. 207 

179. When the irrational parts are binomials. When all 

the irrational parts of the differential expressions are of the 

p 
form (a+ bx) q ; that is, consist of fractional powers of the same 

binomial of the first degree, the expression may be rendered 
rational in terms of z by regarding a+bx as the variable, and 
assuming a + bx=z n , where n represents the least common de- 
nominator of the exponents of the radicals. 

Ex.1. Integrate the expression j. 

(a+bxf 

Assume a+bx—z 2 ^ then, by substitution, we have 

xdx 2{z 2 — d)dz 2z 2 dz 2adz 

(a + ox) 2 

xdx _2z 3 2az__2z 

(a+bxy 1 

2(a + bx) 1F /7 « x „ 

Ex. 2. Integrate the expression 5 j-. 

Assume l+#=s 2 ; then 

dx _ Zzdz 2dz 

(l+xf + {l+xf z*+z~\+z^ 

f * r=2 /ra=2 tang-^2 tang.-(l+^+C. 

_ . xdx 

Ex. 3. Integrate the expression j j. 

(l+xf-(l+xf 
Assume l+x=z 6 . 

xdx (z 6 —l)Qz 5 dz ' . . 
i r =- — ~- 3 — =-6(z 8 +z , +z«+z s +z*+z 3 )dz; 



h 



/xdx n /z 9 z e z 1 z 6 z 6 z i \ _ 

(l +a .)Tr_(i +a .)T \9 8 7 6 5 4/ 



208 INTEGRAL CALCULUS. 

dx 



Ex. 4. Integrate the expression 



(a + bx) 2 1 

A 2(a+bxj* - 
Ans. v b J +G. 

180. When the irrational parts are trinomials. When the 
expression to be integrated contains no radical quantity but 
one of the form Va + bx±Gx 2 , it may be rendered rational by 
substitution. If we remove the factor c from under the rad- 



ical sign, the above expression may be written -\/g \j 



a bx 

-+—±.x 2 , 

G G 

or ^/cVa' ' + b f x±x 2 . It is therefore only necessary to show 
how an expression containing the radical Va+bx±:x 2 may be 
rendered rational. 

Case I. When the sign of x 2 is plus. 

Assume a + bx + x 2 = (z — x) 2 . 

Then a + bx = z 2 — 2zx, 

z 2 — a 
' or x — 7 , Q . 

b + 2z 

Differentiating this value of a?, we obtain 

2(z 2 + bz+a)dz 



dx- 



also, Va + bx + x 2 =z— x — 



(b + 2z) 2 ' 

z 2 + bz + a 



b + 2z 



These values of x, dx, and Va + bx + x 2 , substituted in the given 
differential expression, will furnish a rational expression in 
terms of z and dz, which can therefore be integrated. 

Ex.1. Integrate the expression , =. 

& r Vl+x + x 2 

Assume 1 + x + x 2 = (z — x) 2 . 

m, 2 2 " 1 A j 2(Z 2 + Z+I)dz 

Then ®=p+l' and dx = (2z+l) 2 ? 
/= 1 z 2 +s + l 



IRRATIONAL DIFFERENTIALS. 209 

dx 2(z 2 +z + l) 2z + l . 2dz 

Hence - = , a , ,, n x -^- — rr • dz - 



^l +x+x * (2s+l)2 z*+z + V 23 + 1' 

/" dx . 

J Vl+x+ ^ = ^g.(2z + l)^\og.(2x+l + 2Vl + x+x 2 ) + G. 

Ex. 2. integrate the expression — . 

xVl + x + x 2 

Assume 1 + x + x 2 — (z — a?) 2 . 

<fe 2(s 2 + s + l) 2^ + 1 2z + l 

X 9 -| X 9 -g • $3 



jVl + a+a? - (2z + lf "z*-l"z* + z+l 
2dz dz dz 

= s 2 -l = i^T _ i+l' 



/; 



dfe , 3—1 , x— 1 + Vl + x + x 2 

=r— log. — — ^-=102:. / =; 

#\/i+ff+aj 2 ^+i ^^+i + vi+^+^ 2 



or, multiplying both numerator and denominator by 
#— 1 — Vl+x + x 2 , 
3x 



we obtain log. - 



x 



+ ^ + 2Vl + x + x 2 



181. Case II. When the sign of x 2 is minus. In this case 
the preceding method can not be employed ; for if we assume 
a+bx — x 2 — (z— x) 2 , the second powers of x in the two mem- 
bers of the equation will not cancel each other, and the value 
of x thence deduced will not be rational in terms of z. 
Let j? and q be the two roots of the equation 
x 2 — hx—a = 0; 
then, by Algebra, Art. 434, 

x 2 — bx—a — (x—p)(x — q) ; 
or, changing the signs, 

a+bx— x 2 — {x— p)(q— %). 
Assume V(x—p)(q—x) = (x —p)z. 

Squaring both members, and canceling the common factor 
x—pywe have q—x = (x—p)z 2 . 

Whence x=^ i dx J^~^, 

Z l + \ ' (£ 2 + l) 2 ' 



210 INTEGRAL CALCULUS. 

These values of x, dx, and Va+bx—x 2 , substituted in the 
original expression, will furnish a rational expression in terms 
of z and <fe, which can therefore be integrated. 

Ex. 1. Integrate the expression , =. 

The roots of the equation x 2 — #— 1 = are 

Substituting the values of dx and Vl+x— x 2 before found, 
we have 

dx 2(p—q)zdz z 2 +l 2dz 



Vi+x-x*~ (* 2 +i) 2 {q-j?>~ * 2 +r 

Hence 
f-^==-f^-2 tang.-',=-2 tang.-'x/IES^+C. 



Ex. 2. Integrate the expression —j= 



dx 



Am. -2 tang - ' \/— - + 0. 



x+2 



BINOMIAL DIFFERENTIALS. 211 



CHAPTER III. 

INTEGRATION OF BINOMIAL DIFFERENTIALS. INTEGRATION BY 
PARTS. FORMULAS OF REDUCTION. 

182. Binomial differential defined. A binomial differen- 
tial is a differential expression which contains a binomial fac- 
tor involving the unknown quantity. If the expression be 
rational, its integral can be found by the method of Art. 173. 
We have now to consider the case in which the expression is 

2 _3 _5 P 

irrational, as (3a" 3 " + 5# 4 ) 6 dx, or {ax r + bx n ydx. Such expressions 
can generally be reduced to a simpler form. 

183. How a binomial differential can be reduced to a si?n- 
plerjbrm. 

1. If the variable x were found in both terms of the bino- 
mial, and the expression were of the form 

p 
(ax r +bx n ) g dx y 

we may divide the binomial within the parenthesis by x r (r 
being supposed less than n), and placing this factor without 
the parenthesis, we shall have 

x T {a + bx n ~ r fdx, 
in which but one of the terms within the parenthesis contains 
the variable #, and the exponent of this variable is positive. 

2. If after this reduction the exponent of a?, either within or 
without the parenthesis, or both, should be fractional, we can 
substitute for x another variable with an exponent equal to 
the least common multiple of the denominators of the given 
exponents, by which means the proposed binomial will be 
transformed into one in which the exponents of the variable 
are whole numbers. 



212 INTEGRAL CALCULUS. 

Thus in the example 

if we make x = z 6 , we obtain 

x*(a + bx*y*dx=63 l (a + bz 3 ydz, 
in which the exponents of z are whole numbers. 

Hence every binomial differential can be reduced to the 
form v _ 

x m ~\a+bx n ydx, (1) 

in which m and n are whole numbers, and n is positive. 

184. When this binomial can be integrated. If the expo- 
nent of the parenthesis in binomial (1), Art. 183, is a whole 
number and positive, the expression can be integrated by rais- 
ing the quantity within the parenthesis to the proposed power, 
multiplying each term by the factor without the parenthesis, 
and then integrating each term separately (Art. 160). 

P 

If - is a whole number and negative, we have 

— p x m ~ ^dx 

x m -\a+bx n ) 9 dx=- 



(a+bx n )~ q 

which is a rational fraction, and can be integrated by Arts. 
173-177. 

185. Every binomial differential of the form (1), Art. 183, 

can be integrated ivhen — is a whole number. 
v n 

For this purpose we substitute for the binomial within the 
parenthesis a new variable having an exponent equal to the 
denominator of the exponent of the parenthesis. 

Assume a + bx 11 = z q ; 

then ( a + fa?y=z p . (1) 

Also, x n — — y — , 



and 



(z q — a\ n 



BINOMIAL DIFFERENTIALS. 213 

Differentiating, we have m_ i 

mo /z q —d\* -, 
mx m - l dx = -^z q - l [ — y- J dz. (2) 

Multiplying together equations (1) and (2), and dividing by m } 
we obtain p *_i 

x m -\a + bx n ydx=^ b z p+q -'f^^Y dz. 

77b 

which, according to Art. 184, can be integrated when — is a 
whole number. 

Ex. 1. Integrate the expression 

_3 

du — x 3 (a + bx 2 ) 2 dx. 
Assume a+bx 2 =z 2 . 

Then (a+bx 2 f=z 3 ; (1) 

o z 2 — a ,__ 

also, x 2 — — t — , (2) 

and <vdx=-r-. (3) 

Multiplying together equations (1), (2), and (3), we obtain 



du — x 3 (a + ox 2 ydx = — tt— . dz. 



-n- z 1 az 5 5z 2 —7a 

Hence ^___ = ___ x ^ 

Eeplacing the value of z, we find 

5bx 2 —2a, 7 s _, 
** = 356 2 (a+M 2 +c 

Ex. 2. Integrate the expression du=x 5 (a+bx 2 )*dx. 
Assume a + bx 2 =z 2 . 

Then (a + fe 2 )~*=s, (1) 

**=-gS (2) 

and xdx--r-. (3) 



214 INTEGKAL CALCULUS. 

(z 2 -a\ 2 z 2 dz 
Hence du=l — 7 — ) —r- 

s 6 -2az i +a 2 z 2 
— j3 • d2. 

Therefore 

z 1 2az & a 2 z z 

=^(15^-42^+35^) 



"1056 



=^j^-{W(a+bx 2 ) 2 -±2a(a+bx 2 ) + 35a 2 }+C. 

Ex.3. Integrate the expression du—x 3 (l+x 2 )^dx. 

Ans. u = (1 + a 2 f (8 ^~ 2) + C. 

1 

Ex.4. Integrate the expression du—x 2 {a-\-x^dx. 

3 

Ans. u= V 1Q5 ; (15x 2 -12ax+8a 2 ) + G. 

2 

Ex.5. Integrate the expression du—x & {a-\-bx 2 )~ s dx. 

3{a+bx 2 f*~ 3a(a+bx 2 f 3a 2 (a+bx 2 f 
Ans.u- 22f>3 - 8f>3 + 10 £ 3 - + C. 

_ 1 

Ex. 6. Integrate the expression du=x 3 (a—x 2 ) ~*dx. 

Ans. u= - ^(a - x 2 f{2a + x 2 ) + G. 
Ex. 7. Integrate the expression du—x b {a 2 +x 2 )~ x dx. 
Assume a 2 + x 2 = 2. 

z 2 a* 

u = ^-a 2 z+^ log. z 

/a 2 + x 2 \ 2 a* 

= (— jj— ) - «V + * 2 ) + 2" lo S- (^ 2 + «**) + C ' 



BINOMIAL DIFFERENTIALS. 215 

1 

Ex. 8. Integrate the expression x 3 (a+bx 2 )^dx. 

3bx 2 -2a / 7 3 
Ans. — r^ — (a+ox 2 )*. 

i 
Ex. 9. Integrate the expression x 3 (a 2 +x 2 )*dx. 

3 * 

Ans. -^(a 2 + x 2 f(4:x 2 -3a 2 ). 



Ex. 10. Integrate the expression 



56 v 

Us (XX 

Va + bx 2 

bx 2 -2a / . i 
Ans. — 0T2 — (a + ox 2 ) 2 . 



186. Every binomial differential of the form (1),Jj^.183, 
can be integrated when — + — is a whole number. 

£ 

The binomial x m ~\a + bx n ) q dx may be written 

or ar' 1 (flar n +b) 1[ x T dx 9 

which is equal to x J (ax~ n + b)" q dx, 

which, according to Art. 185, can be integrated when 

np 
m + ^- 
£_ is a whole number, 

n 

or — +— is a whole number. 

n q 

_3 

Ex.1. Integrate the expression du=a(l + x 2 ) ^dx. 
Assume - ar 3 +l=0 2 ; 

that is, 1 + x 2 — v 2 x 2 . 

Then (l + x 2 y^=v- 3 x- 3 -, (1) 

also, * * 

v 2 —l 



216 INTEGRAL CALCULUS. 

,- „ 7 —vdv , M 

Whence dx= x (v 2 -lf ™ 

and l=x\v 2 -l) 2 . (3) 

Multiplying together equations (1), (2) and (3), we have 

3. adv 
du = a(l+x 2 ) ^dx= j-. 

a ax 
Whence w =-=--=== +C. 

v Vl+x 2 x 

Ex.2. Integrate the expression du=x~\l+x 2 )"^dx. 
Assume 1 + x 2 = v 2 x 2 . 

Then flr f =^-l, 

_ 1 

also, #=(y 2 — 1) Tp 

_3 

Whence dx= — {v 2 — 1) T «y^y; (2) 

i l fc 2 _lY* 
also, (1 +^ = _ = L_L. (3 ) 

Multiplying together equations (1), (2), and (3), we have 

du = ar~ 4 (l + % 2 )~ Y dx ——{v 2 — l)dv. 

m v 3 (2x 2 -l)(l+x 2 )^ „ 
Whence u=--^ + v=- ^ -+C 

_ i 
Ex. 3. Integrate the expression ar 4 (l — 2x 2 ) ^dx. 

1 _|_ 4^2 1 

_ 3 

Ex.4. Integrate the expression du—x~ 2 {l + x 2 ) Y dx. 

Ans. - (2a? + - ) (1 + x 2 )~ "* + 0. 

_ i 
Ex. 5. Integrate the expression du=(a 2 + x z ) ^dx. 

i 
Assume ?;— a? + (a 2 + a? 2 ) 2 . 

Then dv = dx + x{a 2 + x 2 )~ ~*dx =—± *-f-dx. 

{a 2 +x 2 y 

Therefore — = (a 2 + x 2 )~~ ^dx. 

Hence u = log. v — log. { x + (a 2 + x 2 Y \ + C. 



INTEGRATION BY PARTS. 217 

2adx 



Ex. 6. Integrate the expression 

xV2ax—x 2 



2V2ax—x 2 



Ans. — 

x 



__ . _ . . xdx 

Ex. 7. Integrate the expression 



(2ax— x 2 \ 2 . ^x 

x J Ans. — , 

av 2a— x 

dx 
Ex. 8. Integrate the expression , 



5 



V¥+ 



,lf 



Ex. 9. Integrate the expression x~ 2 (a + x 3 ) ~*~dx. 

Ans. — — 



3« 3 + 2a 



2a 2 ^(a+^ 3 ) T 

INTEGRATION BY PARTS. 

187. The integral of a differential expression may some- 
times be made to depend upon the integral of another differ- 
ential of a simpler form. This is effected by resolving the 
expression into two parts, one of which has a known integral. 

If u and v are functions of #, then, by Art. 24, 

d(uv) = udv + vdu. 
Whence uv —fudv +fvdu, 

and consequently fudv — uv —/vdu ; (1) 

from which we see that the integral of udv can be obtained, 
provided we are able to integrate vdu. This method of inte- 
grating udv is called integrating by parts. 

To apply this method to a particular example, we resolve the 
given differential into two factors, one of which shall contain 
the differential of the variable, and be capable of integration. 
We substitute this factor for dv in formula (1), we substitute 
its integral for v, and we substitute the other factor for u. 

Ex. 1. Integrate the expression x 3 (l+x 2 ) 4 dx y or x 2 (l + x 2 ) 4 xdx. 

This expression can be resolved into the two factors x 2 and 

K 



218 INTEGRAL CALCULUS. 

(l+x 2 yxdx, the latter of which is the differential of a known 
function. 
Hence we put ^ 

dv = (1 + x 2 )*xdx, whence v = ^(l + x 2 ) 5 ; 

u—x 2 ^ whence du=2xdx. 
Hence fudv =fx 2 (l + x 2 ) 4 xdx = uv —fvdu 



x 2 C 1 

= -(l+^_-(l +aJ 2)6 



'10 v 
"6(A 



=4(l+^ 2 ) 5 (5^-l). 



X uX 3 

Ex. 2. Integrate the expression 3, or x^l+x^'^xdx. 

(i+xY 

This expression can be resolved into the two factors x 2 and 

(1 + x 2 ) ~*xdx, the latter of which can be integrated. 
Hence we put 3 t 

dv=(l + x 2 )~ Y xdx] v— — (l+x 2 )"*; 
u~x 2 \ du=2xdx. 

/TuX 
3 — uv —fvdu 

^ J ( 



(i+%y j (i+x*y 



- + 2(1+0' 



1 

,2Y2T 



(i+xy 

x 2 + 2 

{l + xrf 
Ex. 3. Integrate the expression x log. xdx. 

Put dv—xdx^ v = -^, 

u=\og.x, du— — . 
x 






INTEGRATION BY PARTS. 219 

Hence fx log. xdx — tcv —fvdu 

.. X i X ctx 

& 2 J 2 x 

X 2 X 2 

=2 log. a?- ^ 

=|(log.*-|). 

_ T . . a? 3 6fe 3# 2 + 2 
Ex. 4. Integrate the expression ^. J_/is. — 5. 

(l + x 2 f 3(1 +a? 2 )* 

Ex. 5. Integrate the expression af log. xdx. 

Put w = log. x, and cfo = af <fe. 

„ . , x n+l \og.x Cx n dx 

x n io^.xax~ -f — — / — —r 

J & n + 1 J n + 1 

X n+1 loff. X x n+1 



n+1 (n + 1) 2 ' 

* JL 

Ex. 6. Integrate the expression x 3 (a—x 2 ) 2 dx. 

Put u—x 2 , and dv=x(a—x 2 ) 2 dx. 

fx\a—x 2 ) 2 dx= — -w(a—x 2 ) 2 + I - — *-*—< 
x 2 3. 2 A 

= _- _(tf_£2)2_ /^_^ 2> 



2xdx 



1 

_ „ T , . (1 — x 2 )^dx 
Ex. 7. Integrate the expression ~ . 

Put u = (1 — a? 2 )" 5 ", and dfo = -g . 

■ 

«(l-a? 2 )*<fo (1-^ /• dx 



m-a?)*dx (l-x 2 Y r 
J x 2 x ~ J 



1 

\2 



8 

Ex. 8. Integrate the expression (a 2 —x 2 ) 2 dx, 

1 
Put u — {ct 2 — x 2 ) 2 y and dv = dx. 



(l-x 2 f 

(l — x 2 Y 

-sin r l x (see Art. 168). 



220 INTEGRAL CALCULUS. 

1 l /• X 2 dx 

f{a 2 -x 2 fdx=x{a 2 -x 2 Y+ / — (1) 

J {a 2 -x 2 f 

But f(a 2 -x 2 fdx= / — t*b 

■J (a 2 -a; 2 ) 9 

r- f i- (2) 

(a 2 -x 2 ) 2 J (a 2 -x 2 ) 2 

Adding equations (1) and (2), we have 

2/(a a -aj 2 )*<fo=aj(a a -«*)*+ a 2 / — T . 

J (a 2 -xy 

Hence f(a 2 —x 2 )^dx=:^(a 2 —x 2 )^+-^ sin. -1 - (see Art. 168). 

188. By the application of the preceding method many 
binomial differentials can be integrated. For greater con- 
venience, we shall henceforth represent any binomial differen- 
tial by the form x m (a + bx n ) p dx, 

where p is supposed to be a fraction, but m and n are whole 
numbers, and n is positive. 

METHOD OF REDUCTION. 

189. This method consists in making the integral of the 
proposed expression depend upon the integral of another dif- 
ferential of the same form, but in which the exponents are 
diminished, so that by repeating the process we at length ob- 
tain a differential expression which can be integrated by meth- 
ods already established ; or, in case integration is not possible, 
we have the part not integrated in the simplest form possible. 

190. The integral of any differential of the form 

x m (a + bx n ) p dx 
may be made to depend upon the integral of 

x m ~ n (a + bx n y+ x dx. 
Assume dv = (a + bx n ) p x n ~ l dx; when ce u — x m ~ n+1 . 



METHOD OF REDUCTION. 221 

Integrating the former expression, and differentiating the latter, 

we have (a + bx n ) p+l 

V= nb(jp + l)> 

and dii — {m—n + l )x m ~ n dx. 

Substituting these values in the formula fudv — uv—fvdu, we 

have 

fx m (a+bx n ) p dx 

x m - n+l (a+bx n ) p+l m-n + 1 . , . N ^ 17 
= /) , ix - -77 — 7^\M~ n (a + bx n Y +x dx. (1) 

Equation (1) is called a formula of reduction, and by means 

of it we make the integral of x m (a + bx n ) p depend on that of 

x m ~ n (a + bx n ) p+ \ In the same way the latter integral can be 

made to depend on that of x m ~ 2n (a + bx n ) p+ % and so on. For 

3 
example, if m — 2, n — 2, and^?= — ^, we have 

/x 2 dx x 1 r dx 

(a + bx 2 y bVa + bx 2 bJ Va + bx 2 
The integral of the latter expression can be found, and hence 
the integral of the former expression becomes known. 

191. The integral of any differential of the form 

x\a + bx n ) p dx 

may be made to depend upon the integral of 

x m +\a + bx n ) p - l dx. 

From Equation (1), Art. 190, by transposition, we obtain 

\ -, , ^ , x m - n+ \a+bx n ) p+l nb(p+l) n , 7 N 7 
fx m - n (a+bx n ) p+l dx= * - T l — -- w . J Jx m (a+bx n ) p dx. 

For greater convenience, substitute m for m—n, and j? for 
jp + 1, and we have 

fxHa + bx^dxJ^-^ (2) 

j \ j m + 1 m + l J v ' x J 

By means of Eq. (2) we make the integral of x m (a + bx n ) p dx 
depend on that of af*+*(a+5af Y~ l dx. In the same way the latter 
integral can be made to depend on that of x m+in (a + bx n ) p ~ 2 dx; 



222 INTEGRAL CALCULUS. 

and by repeated applications of the formula the exponent of 
the binomial, if positive, may be reduced to a fraction less than 
unity, either positive or negative. 

192. Modification of formulas (1) and (2). Equations (1) 
and (2), Arts. 190 and 191, may be reduced to forms which 
are more convenient for use. 

Since 

(a+bx n y +1 = (a+bx n ) p (a+bx n )=a(a+bx n y+bx n (a+bx n y, 
if we multiply by x m ~ n dx, we shall obtain 

ar-*(a + bx n ) p+1 dx = ax m ~\a + bx n ) p dx + bx m (a + bx n ) p dx. 

Substitute this value in the last term of equation (1), Art. 
190, multiply the result by nb(p + l)> transpose the last term 
of the product to the first member of the equation, and divide 
the equation thus obtained by b(njp + m + 1), and we shall obtain 

Formula A. 

^-^Ua^lxy^-aim-n^fx^Ha+bxydx 

/ar(a+tefy<te= i b<M+ m +i) — • 

By means of Formula A the integral of the proposed differ- 
ential is made to depend upon the integral of another differen- 
tial in which the exponent of the factor x m without the paren- 
thesis is diminished by that of <z n , the variable within the 
parenthesis. By a second application of the same formula, 
we should find ' fx m -\a + bx n ) p dx 

to depend upon fx m -* n {a + bx n ) p dx ; 

and, by continuing the process, the exponent of the factor with- 
out the parenthesis may be diminished until it is less than n. 
If m is positive, and if ?n + l is an exact multiple of n, then, 
after a certain number of applications of the formula, the in- 
tegration will be entirely accomplished. 

193. When the exponent m is negative. By the application 
of Formula A the exponent m will be numerically diminished 
only when m is positive ; but from this formula we may de- 
duce another, which will numerically diminish the exponent 



METHOD OF REDUCTION. 223 

when m is negative. For this purpose, clear Formula A of 
fractions, transpose the term which does not contain the sign 
of integration, divide by a(m—n+l\ and we obtain 

fx m - n (a+bx n ydx= ^ 1 /^ )J v 1 — . 

For greater convenience, substitute m + n for m, and we obtain 

Formula B. 

, 7 % 7 x m+ \a+bxy +l -b(np^m^n+l)fx m+n (a^bx n ) p dx 

f x\a+bxydx=-± 1 KJ <m +i) — : 

By the application of this formula, when m is negative, m 
will be numerically diminished by the number of units in n. 

Q7h —I— 1 

If np + m + n + 1 = 0, or ?i= — — — j, the part to be integrated 
will disappear, and the integration will be complete. 

194. To diminish the exponent of the parenthesis. 

Since 
(a + bxy=(a+bxy-\a+bx n )=a(a+bx n y- 1 + bx n (a+bx n ) p -\ 
multiplying each member by x m dx, we obtain 

x\a + bx n ) p dx = ax m (a + bx n y~\lx + bx m+ \a + bx n ) p ~ l dx ; 
and integrating, we have 

far (a + bx n ) p dx = q/2zr(a + bx n ) p ~ l dx + bfx m+n (a + bx n ) p ~'dx. 
Transposing the terms of Equation (2), Art. 191, we obtain 

fx m+ Xa+bx n ) p - l dx= ^ * ^—fx m (a+bx n ) p dx. 

Hence, by substitution, we have 

fx m (a+bx n ) p dx 

= a/a^a + bx n y~ l dx + v — ^-— — - 1 — /affa + baTYdx. 
Transposing and uniting terms, we obtain 

Formula C. 

/affo + &» n V <?# = ^— I ; , ) 

J v y np + m + 1 

By this formula the value of the required integral is made 



224 INTEGRAL CALCULUS. 

to depend upon another in which the exponent of the binomial 
is diminished by unity. By a second application of the same 
formula the value of this new integral may be made to depend 
upon that of an integral in which the exponent of the binomial 
is still further diminished ; and by repeated applications of the 
formula the exponent of the binomial, if positive, may be re- 
duced to a fraction less than unity, either positive or negative. 

195. When the exponent p is negative. By the application of 
Formula C the exponent^ will be numerically diminished only 
when^> is positive; but from this formula we may deduce 
another, which will numerically diminish the exponent when 
j> is negative. For this purpose, clear Formula C of fractions, 
transpose the term which does not contain the sign of integra- 
tion, divide by anp, and we have 

J x ; a?ip 

For greater convenience, substitute p + 1 for jp, and we obtain 

Formula D. 

fx-(a+bxydx= * anfr + 1) 

By the application of this formula, when p is negative, jp 
will be numerically diminished by unity. Ii_p = — 1, the sec- 
ond member of the formula becomes infinite ; but in this case 
the primitive expression is a rational fraction, and may be in- 

n% -\- 1 
tegrated by Arts. 173-177. If nj?+n+m+l=0, or n— — ^ 

the part to be integrated will disappear, and the integration 
will be complete. 

196. Modification of Formula A. We have frequent occa- 
sion to integrate binomial differentials of the form 

x m dx 

V2ax—x 2 



METHOD OF REDUCTION. 225 

If we remove the factor x from under the radical sign, this 

l _x_ 

expression may be written x" *(2a—x) 2 dx, which may be in- 
tegrated by Formula A, by making the proper substitutions. 
Substitute m— \ for ra, 2a for a, — 1 for &, 1 for n y and — £ 
for j9, and we have 3 i 

/# H2a-x) ^dx= : . 

or, changing the form, w r e derive 

FORMULA E. 



/x m dx x m ~W^ax—x 2 a(2ra— 1) C 



V2ax-x 2 ™ m J Vlax—x 2 

By the application of this formula the exponent ?n is dimin- 
ished by unity. If m is entire and positive, by repeated appli- 
cations of the formula we ultimately arrive at the form 

dx 



h 



Vlax—x 2 * 

which is equal to the arc of a circle whose versed sine is —(see 
Art. 170). 

197. Application of the preceding formulas. By the appli- 
cation of Formulas A, B, C, D, and E, a great variety of differ- 
ential expressions may be integrated. 

Ex.1. Integrate the expression x 3 (a + bx 2 )~ ~*dx. 

In Formula A substitute 3 for m, 2 for n, and — -j for p, 
and w r e obtain x t 

. 3 , . 2 . . i x\a + lx 2 ) a - 2afx(a + bx 2 )~ *dx 
fx 3 (a + ox 2 ) "dx— wt ■ . 

By a second application of the formula the integral of 

x{a-\-bx 2 )~^dx is found to be 7 — -. 

Hence we have 

f x 3 dx _(& 2a\ 

J VaTW-\3b-3b*)^ a +!> x2 + V- 

K2 



226 INTEGRAL CALCULUS. 

Ex. 2. Integrate the expression x 5 (a + bx 2 ) 2 dx. 
By three successive applications of Formula A we obtain 

/x* 4:ax 2 8a 2 \, _ n i ^ 

_ i 
Ex. 3. Integrate the expression x~\a+bx 2 ) ^dx. 

In Formula B substitute —4 for m, 2 for n, and — -| for jp, 

and we obtain x t 

/a? 4 (# + te 2 ) 2 aa?= ^k — . 

By a second application of the formula the integral of 

x~ 2 (a+bx 2 )~ 2 dx is found to be — . 

v ' ax 

Hence we have / , = (— o — 3 + o~r") Va+JaJ 2 + 0. 

•/ a*Va + bx 2 \ 3ax 3 ^3a 2 x/ 



Ex.4. Integrate the expression x~ 6 (a + bx 2 ) ^dx. 
By three successive applications of Formula B we obtain 
. / 1 4ft 8b 2 \ , — ^ 

i 
Ex. 5. Integrate the expression (a+# 2 ) T ^#. 

In Formula C substitute for m, 1 for b, 2 for w, and \ for 

j?, and we obtain x 

/$ x 

, =log. {# + Va + x 2 }. 

Ya+x 2 

Hence we have x 

f(a+x 2 fidx= X ( a+ 2 X ^ + | log. {x+{a+x 2 f} + 0. 

8 

Ex. 6. Integrate the expression (a + x 2 J*dx. 

By Formula C, 3 

ox-3- -r a?(# + # 2 V* 3& _, n i _ 
Aa + xydx= v ^ ; + T /(a+* 2 )^. 

Hence, by Ex. 5, 8 

m /{a+x 2 )*dx= v 4 y +-g-(a+a 2 ) 7 +-g- log.jtf+^+a! 2 )^} +0. 



METHOD OF REDUCTION. 227 



3 



Ex. 7. Integrate the expression (a + bx 2 ) *dx. 
Substituting in Formula D, we obtain 

f{a + bx 2 )~ ^dx=-(a+bx 2 )~^+G. 

Ex. 8. Integrate the expression (a 2 + x 2 )~ 2 dx. 
By Formula D, 

9X o7 x(a?+x 2 Y l 1 r dx 



a 2 + x 2 ' 

/dx x 

— — z = - x the arc whose tangent is -. 
Cl -\- X d CO 

Hence /fo a +^-'^= 2gV+*») + 2g **"*5+ a 

_ 1 

Ex. 9. Integrate the expression x 2 (2ax— x 2 ) ^dx. 
By two applications of Formula E we obtain 

x+3a.^ n i 3& 2 . ,o? _, 

Ans. — — 5 — (2ax— x 2 y-\ — <r- ver. sm ~ 1 -+0. 

3. _ i 

Ex.10. Integrate the expression x 2 (2ax— x 2 ) ^dx. 

8a + 2x,^ i „ 

jiw*. - — g — (2a-#) 2 + C. 

Ex.11. Integrate the expression x\a+x 2 ) 2 dx. 

{ x 3 3ax ) , - 3a 2 . / , _. 

Ans. < -T—-Q- \ Va+x 2 +-glog. (x+Va+x 2 ) + G. 

5 

Ex.12. Integrate the expression x 3 (a + hx 2 ) 2 ~dx. 

Ex.13. Integrate the expression x'{a-\-bx 2 ) ' J dx. 

[ x e Sax* 8a 2 x 2 16a 3 ) . , a _ 
Ans - \ 76-35^+ W8W 1 f (* + ^ T +°- 

_ 3 

Ex. 14. Integrate the expression aj 3 (a + Ja? 2 ) ^dx. 

{ x 2 2a) , 7 ox _ i ~ 



228 INTEGEAL CALCULUS. 

_5 

Ex.15. Integrate the expression (a+bx 2 ) ^dx. 

. x 2x 

Ans. 3+ T + U- 

3a(a + bx 2 f 3a 2 (a + bx 2 f 

Ex.16. Integrate the expression x*(l— x 2 ) 2 dx. 

ix 3 Sx) ,- 3 . _ ._ 

Ans. — < t+"q~ f vl-r + o sm.-^+C. 

_ 3 

Ex.17. Integrate the expression x~ 2 (a+bx 2 ) ^dx. 

(1 2bx) , 7 x i _ 

_ 7 

Ex.18. Integrate the expression x 3 (a+bx 2 ) ~*dx. 

[ x 2 2a ) , _ * 
Ans. - J gj+jgga [ (^+^ 2 ) ^+C. 

Ex.19. Integrate the expression x~\a+bx 2 y*dx. 3 

Ex. 20. Integrate the expression x 3 (a+bx 2 ) 1I dx. 

Ans. | j^-ggfi I (a + fo 8 )*+C. 

Either of the Formulas A, B, C, D, and E may become inap- 
plicable on account of the denominator vanishing for special 
values of m, n, and p ; but in such a case the proposed differ- 
ential expression may be rendered rational, and its integral 
may be found by methods already explained. 



INTEGRATION BY INFINITE SERIES. 229 



CHAPTER IV. 

INTEGRATION BY INFINITE SERIES. 

EXPONENTIAL, AND TRIGONOMETRICAL DIFFERENTIALS. 

198. When a proposed differential expression can not be 
integrated by any of the ordinary methods, we may expand it 
into an infinite series, the terms of which can be integrated 
separately. If the result is a converging series, the approxi- 
mate value of the integral may be determined by summing a 
finite number of terms. This method is also sometimes useful 
in developing a function into a series. 

dx 
Ex.1. Integrate the expression r— , or (l+x)~ l dx. 

By the binomial theorem, we have 

(l+x)~ 1 = l—x + x 2 —x 3 -\- etc. 
Multiplying by dx, we have 

(1 + x)~ l dx =dx— xdx + x 2 dx — x 3 dx + etc. ; 
and integrating each term separately, we obtain 

rpi /v»3 rjn& 

f(l+x)- l dx=x- g- +3— j + . . . +0. 
But /_^_ = i og .(i +4 

sy)2 /j)3 ryA 

Hence log. (l + x)=x— •'cr+ir— "T+ e ^ c - 

dx 

Ex. 2. Integrate the expression -, 2 * n an infinite series, 

J. - f- X 

and thus obtain a development of tang. _1 #. 
By the binomial theorem, we have 

(l + x 2 )-* = l-x 2 + x 4 — x 6 + etc. 

Hence (l + x 2 )~ 1 dx=dx— x 2 dx + x*dx—x 6 dx+ etc. 

m, , f dx x 3 x 5 _. 

Therefore tang. _1 ^= / 1 2 — x— -»+t— . . . +C. 



230 INTEGEAL CALCULUS. 

But when a?=0, tang.^a^tang.^O^O. Therefore C = 0, and 

ry%& /v>0 /yj' /yi9 

tang" 1 a?=aj— g + g-— y+-g— etc. 

dx 
Ex. 3. Integrate the expression in an infinite series, 

VI— x 2 

and thus obtain a development of sin. _1 a?. 

x 3 1.3a* 1.3.5z 7 
Ans. x+ — + ^^ + ^-^ + etc. 

dx 

Ex. 4. Integrate the expression — — in an infinite series, and 

thus obtain a development of log. (a + x). 



>,2 



rjn qnC, rpo qnf* 

Ans. \og.(a+x)=log. a +a~2tf + 3tf~Itf + etc ' 

i i 

Ex. 5. Integrate the expression x*(l—x 2 J*dx. 

A JL it i s 
2a? 2 as 2 a? 2 aT 2 " 

i 
Ex. 6. Integrate the expression x 2 (l—x 2 ) Y dx. 

•yid /yiO /yW /yj9 

. tLf %K/ XK/ lis _. 

Ans. 3 -io-66-iii- •••+<* 

INTEGEATION OF LOGARITHMIC DIFFERENTIALS. 

199. Logarithmic differentials do not generally admit of 
jexact integrals, but when the exact integral can not be ob- 
tained, an approximate integral may be obtained by infinite 
series. We will examine a few of the simplest forms of log- 
arithmic differentials. 

Ex. 1. Integrate the expression x m log. xdx. 

Assume u == log. x, and dv = x m dx ; 

dx x m ^ 

which give du = — , and v = 7. 

to x ' m + 1 

Hence, Art. 187, we obtain 

„ . , x m+1 loo;, x C % m dx 

fx m lo^.xdx— -? — — / — -T7 

J & m + 1 J ra + 1 

# m+1 ( , 1 



"m + 1 



| 1o s^-^tt}+ c - 



INTEGRATION OF LOGARITHMIC DIFFERENTIALS. 231 

Ex. 2. Integrate the expression a? 4 log. xdx. 

x 5 ( 1 ) 

Ans. y jlog.B-g J +0. 

Ex. 3. Integrate the expression af(log. x) 2 dx. 

Assume (log. xf — u, and x m dx = do ; 

_ . , . - , 2 loo;, xdx . af +1 
which 2*1 ve du — - , and v = — nr- 

& x m + 1 

Hence, Art. 187, we obtain 

x m+l (log x) 2 2 

fx m (\02. Xfdx = , ^ ^ -r /bf 1 log. £^. 

^ \ & j m + 1 ra + 1* 7 & 

Integrating the last term by Ex. 1, we have 

fa r{\o S .xfdx=^ l | flog-^-^^+^p j +c. 

Ex. 4. Integrate the expression # 2 (log. xfdx. 

x 3 I „ N 2 loff. x 2 ) _, 
-An*. 3- j (log.^) 2 -— ^— +9 J +C. 

Ex. 5. Integrate the expression af (log. x) n dx. 

Assume (log. x) n = u, and x m dx — dv ; 

dx x m+l 

which give du — ?i(log. x) n ~ l — , and v = -. . 

Hence, by Art. 187, we obtain 

••*■/! nt af l+1 0oo;. sc)* ^ . _ . . . 
/^(log. ac)»<fa= m + 1 -^+i/ tfW ( 1 °g- *) ^ 

This formula diminishes the exponent of log. x by unity, 
and by successive applications of the formula to the last term 
of the expression, the exact integral will at length be found 
by means of Ex. 1, provided n be a positive whole number. 

Ex. 6. Integrate the expression af (log. x) 3 dx. 

. x 6 ( (\o°;.x) 2 loo\# 1 ) _, 

200. TFAm m in Ex. 5, ^Lr£. 199, is equal to -1. The for- 
mula in Ex. 5 fails when •»=— 1, for the second term of the 
expression becomes infinite. The integration may, however, 
be effected as follows : 



232 INTEGRAL CALCULUS. 

Assume log.x=z. 

Then — — dz\ 

X ' 

dx „ _ s n+1 (log.#) w+1 

/(log. #) n — =/s"^fo = — -7 = V 6 , ; . 

t/x to J x J n+1 n + 1 

Ex. 1. Integrate the expression x~ l log. xdx. „ , 2 

Ans. {0 & X) + G. 

Ex.2. Integrate the expression ar^log. xfdx. 

(log. xf _ 

J.7W. V ° ^ +0. 



201. TF/^?i n in Ex. 5, J.Ttf. 199, is negative. The method 
given in Ex. 5, Art. 199, fails when the exponent n is negative, 
since, by application of the formula, the exponent of log. x y 
instead of approaching zero, continually becomes a greater 
negative number. In this case we may put the proposed dif- 
ferential under the form 

dx 

af l+1 (log. x)~ n — , or af l+1 (log. x)~ n d log. x. 

Assume u =af" +1 3 and 6?^ = (log. #)~ n <# log. x ; 

(log. #)-" +1 



which give du — (m + l)x m dx, and : 



-71+1 ' 



fx m (\og. x)~ n dx= —, ^vtt r^- 

* * » ' (?i— l)(log. X) 



fir afvfo 

l/ (log.^- 1 ' 



If n be a whole number, the repeated application of this 

x"'dx 
formula will at length give for the term to be integrated ; , 

In order to simplify this expression, put z=x m + l . °* 

los*. z 
Whence (m + 1) log. #=log. 2, or log. x= _, . 

Also, by differentiating, 

(in + V)x m dx — dz t 

Whence x m dx-- 



Thcrefore 



??i + l 



log. a? log. s' 
a differential which can only be integrated by series. 



INTEGRATION OF EXPONENTIAL DIFFERENTIALS. 233 

Ex.1. Integrate the expression x i (\og.x)~ 2 dx. 
By the preceding formula, 

/ x 4 dx x 5 Cx^dx 

(log. a?) 2 "" log. x J log. x 

x^dx 

Ex. 2. Integrate the expression 7 , r~. 

G r (log. xy 

x 5 5x 5 25 Cx*dx 

~~ 2 (log. x) 2 ~~~ 2 log. x2J log. x 

202. By similar methods we may integrate X(log. x) n dx, in 

which X denotes any rational function of x. 

loo*, acdx 
Ex. 1. Integrate the expression ,r^_ 2 . 

Assume u — log. a?, and dv = (l— x)~ 2 dx ; 

dx 
which give <#w = — , and ^ = (1— a?)"" 1 . 

x 

Hence, Art. 187, we obtain 

/log.xdx log. a? /* <fe 
(l-^) 2 = ll^""J a?(l-a>) 

log. x rdx r dx 

~~ 1— X J X J 1 



a* 



# log;, a? . , H . _ 
= T ^-+log.(l-B) + C. 

Ex.2. Integrate the expression (a +5a?) wl log. xdx. 



Ans. 



{a + bx) m+l log. x 1 /• (a + bx) m+l dx 



(m + l)b (m + l)bJ x 



LA 



INTEGEATION OF EXPONENTIAL DIFFERENTIALS. 

203. Exponential differentials are closely related to loga- 
rithmic differentials, and do not generally admit of exact integ- 
rals ; but we can usually cause the value of the required in- 
tegral to depend upon another which is of a simpler form. 
We will consider a few of the simplest cases. 

Ex.1. Integrate the expression xa x dx. 



234 INTEGRAL CALCULUS. 

Assume u = x, and dv = &*<& ; 

which give du — dx^ and ^^r - ^ — (see Art. 40). 



TT _ xa x C a x dx 

Hence fxa x dx=, — / n 

J log. a J log. a 

xa x a x 

"log. a (log. of 

Ex. 2. Integrate the expression x m a x dx. 

Assume u = af , and $y = a"^» ; 

which give du = mx m - J dx, and t>: 



ar 



Hen ce fx m a x dx = , — , fx m ~ ] a'dx. 

J log. a log. or 

By successive applications of this formula the exponent of 

x, if positive, will be continually diminished, and if it be also 

a whole number, the exact integral may be determined. If 

the exponent of x be a fraction, the integral can only be found 

in an infinite series. 

Ex. 3. Integrate the expression x 3 a x dx. 

x ( x 3 3x 2 6x 6 

( log. a (log. a) 1 (log. a) 3 (log. ay 

204. When the exponent m in Ex. 2, Art. 203, is negative. 
The method given in Ex. 2, Art. 203, fails when the exponent 
m is negative, since, by successive applications of the formula, 
the exponent of x, instead of approaching zero, is increased 
numerically. In this case we may proceed as follows : 

Assume u — a x , and dv = x~ m dx ; 

which give du — a x log. adx, and v— — -, 1X m _ v 

[771 — _L jX 

__ Ca x dx a x log. a Ca x dx 

Hence / — — — —1 ix m -i + c 1 / ~^3t- 

J x (?n—l)x m 1 m — U x m l 



(?n—l)x' 
Ex. Integrate the expression 



a x a x log. a (log. a) 2 
AnS - ~2^~ 2x + ~ '~2~ 



1 C a x dx 
J x 



INTEGKATION OF EXPONENTIAL DIFFERENTIALS. 235 

nfdx 

The expression — — can only be integrated by development 
into series, as shown in the following Article. 

205. When the exponent m in Art 204 is equal to unity. 
The formula in Art. 204 fails when the exponent m is equal 
to unity, since the denominator of the second member becomes 
zero. In this case the integral can only be determined by de- 
veloping a x into an infinite series, and integrating each term 
separately. According to Ex. 12, Art. 56, we have 

a x =l+x log. a + -gGog. a) 2 + ^(log. df + etc. 

a x dx dx X X 2 

Hence = — +log. adx + ^Q-Og. a) 2 dx+^- o(log. a) 3 dx+ etc. 

XX & Zi » o 

Therefore 

/a x dx _ x 2 - xo x 3 n .. 

-^- = log. x + x log. a + -j (log. a) 2 + jg(log. ay + etc. 

This formula solves the exceptional case mentioned in Art. 204, 
and shows the intimate relation between logarithmic and ex- 
ponential integrals. 

206. By a similar method we may integrate the expression 
X.a x dx, in which X represents any algebraic function of x. 

Assume u — X, and dv = a x dx. 

Now da x = a x log. adx ; 

. . 7 da x 

that is, a x dx — , , 

log. a/ 

and fa x dx— 1 —v. 

J log. a 

Xa x 1 

Hence fX.a x dx = ? — . fa*d X. (1) 

J log. a log. or v J 

By the application of this formula the integral of 3La x dx is 
made to depend upon the integral of a x dX. If the function 
X is of such a form that one of its differential coefficients is 
constant, the next differential coefficient will be zero. In this 
case the exact integral may be obtained by successive applica- 
tions of formula (1). 



236 LNTEGKAL CALCULUS. 

207. When the constant (a) in Art. 203 is the hase of the sys- 
tem of logarithms. If we make the constant a in Art. 203 
equal to the base of the Naperian system of logarithms, then, 
since the logarithm of the base is unity, the preceding results 
will assume a simpler form. In the following examples the 
base of the Naperian system is represented by e. 

Ex.1. Integrate the expression xe x dx. Ans. e x (x— 1) + C. 
Ex. 2. Integrate the expression x 2 e x dx. 

Ans. e x (x 2 -2x+2) + C. 
Ex. 3. Integrate the expression x 3 e x dx. 

Ans. e x (x 3 -3x 2 + 6x-6) + C. 
Ex. 4. Integrate the expression x*e x dx. 

Ans. e x (x*-4:x 3 + l2x 2 -24:X+24:) + C. 

Ex. 5. Integrate the expression . 

Ans. log. %+%+o2+k- Q2+ etc. 

_ _ _ . e x dx 

Ex. 6. Integrate the expression — %-. 

e x _ x 2 x 3 

Ans. --+log.x+x+^ 2 +jjp+ etc. 

Ex. 7. Integrate the expression — —. 

e 

Assume u = as, and dv — e~ x dx ; 



r> — * 



which give du — dx^ and v=—e 

Cxdx fdx 

Hence J — = - e -.x+J v 

= —e—.x—6—+C. 

x 2 dx 
Ex. 8. Integrate the expression — — . 

e 

Assume %b — x 2 , and dv = e~~ *dx ; 

which give du = 2xdx, and v= — e~ x . 

2xdx 



fx 2 dx \ r 

Hence / — — = — e~ x .x 2 + J 



= -e- x (x 2 + 2x+2) + C. 



INTEGRATION OF TRIGONOMETRICAL DIFFERENTIALS. 237 



INTEGRATION OF TRIGONOMETRICAL DIFFERENTIALS. 

208. Trigonometrical differentials may sometimes be reduced 
to algebraic forms by the substitution of a new variable, and, 
when this is impossible, they may sometimes be integrated by 
methods similar to those employed in the preceding Articles. 

Ex. Integrate the expression (sin. x) m dx. 
Assume sin. x — z. 

Whence cos. x = (l — z 2 y ; 

dz * 

dx — — Ci—z 2 ) 2 dz. 

COS. X x J 

Hence /(sin. x) m dx =fz m (l — z 2 )~ ^dz. 

This expression can be integrated by the method of Arts. 
192 and 193. When m is a whole number, by repeated appli- 
cations of Formula A or B the integral may be made to depend 

/dz . . /• zdz 
f, which is equal to #, or upon / j, which 
(1-z 2 )* J (1-s*) 7 
is equal to —cos. a?. 

In particular cases simpler methods of integration are pos- 
sible. 

209. When the exponent m in Art. 208 is an odd positive 
integer. 

Suppose m = 2r + l. 

Since (sin. x) m = (sin. 2 #) r sin. x = (1 — cos. V) r sin. a?, 
we have /(sin. x) m dx =/(l — cos. 2 #) r sin. xdx 

— —f(l — cos 2 x) r d cos. x. 
Since 7 1 is an integer, this binomial can be expanded in a finite 
number of terms, and each term can then be integrated sepa- 
rately. If m is an odd negative integer, this substitution gives 
a rational fraction in cos. x which is integrable. 

Ex. 1. Integrate the expression (sin. xfdx. 3 

Ans. —f(l — cos 2 x)d cos. x— —cos. x + — ^ — . 

o 

Ex. 2. Integrate the expression (sin. x) 5 dx. o B 

A *= 9 * 7 2 cos. 3 # cos. 5 # 

.Ans. — J (1 — cos. 2 e»)^ cos. x — — cos. a? + — 5 — — — = — . 



238 INTEGKAL CALCULUS. 

210. When the exponent m in Art. 208 is an even positive 
integer. In this case we may develop the power of sin. x in a 
series of terms of sines or cosines of multiples of the arc #, and 
then integrate each term separately. 

By Trig., Art. 74, 2(sin. x) 2 = l- cos. 2x. 
By involution and substitution, we obtain 

8(sin.^) 4 = 3— 4 cos. 2x + cos. 4#; 
32(sin. xf = 10 — 15 cos. 2x + 6 cos. 4# —cos. 6x. 
Ex.1. Integrate the expression (sin. xfdx. 

A n /. r, 7 ^^ S ^ n - ^X X „ 

Ans. —^fcos.2xdx+ I "o=— — r — + o + C. 

Ex. 2. Integrate the expression (sin. xfdx. 

. sin. 4a? sin. 2x Sx _. 
-Aw. -7^ 4- + "8 +C - 

Ex. 3. Integrate the expression (sin. xfdx. 

sin. 6# 3 sin. 4# 15 sin. 2x 5x 
Ans. --392"+ 64— 64 + l6 + °' 

211. It is required to integrate the expression (cos. xfdx. 

Assume cos. x — z. 

i 
Whence sin. x = (1 — z 2 f*. 

Hence /(cos. #) n ^ = -fz n (l - a 2 )"" 7 <fe, 

which expression may be integrated as in Art. 208. 

When n is an odd positive or negative integer, we may pro- 
ceed as in Art. 209, and will have 

/(cos. xfdx=f(l—sm. 2 xfd sin. x, 
where the binomial may be expanded, and each term integrated 
separately. 

Ex. 1. Integrate the expression (cos. xfdx. 

(sin. a:) 3 
Ans. sin. x— — „* — +(J. 

Ex. 2. Integrate the expression (cos. xfdx. 

2(sin. xf (sin.cc) 5 
Ans. sin. as q + — g — +^- 

When n is an even positive integer, we may develop the 






INTEGRATION OF TRIGONOMETRICAL DIFFERENTIALS. 239 

power of cos. x in a series of terms of cosines of multiples of 
the arc a?, and then integrate each term separately. 

By Trig., Art. 74, 2(cos. x) 2 = 1 + cos. 2x. 
By involution and substitution, we obtain 

8 (cos. a?) 4 = 3 + 4 cos. 2x + cos. 4a? ; 

32(cos. a?) 6 = 10 + 15 cos. 2x + 6 cos. 4a? + cos. 6x. 

Ex. 3. Integrate the expression (cos. x) 2 dx. x g j n % x 

Ans. 2+— j— . 

Ex. 4. Integrate the expression (cos. a?) 4 <ia?. 

, 3a? sin. 2a? sin. 4a? 
^..-g-+- T -+-32-. 

Ex. 5. Integrate the expression (cos. x) 6 dx. 

5x 15 sin. 2x 3 sin. 4a? sin. 6a? 
^- /l * 16 + 64 + ~~ 64 + "192~' 

212. It is required to integrate the expression 

(sin. a?) m (cos. x) n dx. 

Assume sin. x — z. 

i 
Whence cos. x = (1 — z 2 ) 2 ; 

Hence /(sin. #) w (cos. x) n dx=fz m (l-z 2 ) 2 (l-z 2 y~*dz 

n— 1 

==/ar(li-^)"T^ 

which may be reduced by Formulas A, B, C, and D, and in 
some cases integrated. 

When either m or n is odd and integral, whether positive or 
negative, it is more convenient to follow the method of Art. 
209. If m = 2r + 1, we shall have 

/(sin. a?) w (cos. x) n dx— —/(cos. x) n (l — co$ 2 x) r d cos. a?, 
which may be expanded and integrated as in Art. 209. 

If n=2r + l, we shall have 

/(sin. a?) m (cos. a?) w 6?a?=/*(sin. a?) w (l — sin. 2 a?) r ^ sin. a?. 
If both m and ^ are positive and even integers, substitute 
2 sin. 2 a? = 1 — cos. 2x and 2 cos. 2 a? =1 + cos. 2x, expand and reduce 
by Art. 211. 



240 INTEGRAL CALCULUS. 

Ex. 1. Integrate the expression sm. 3 x cos. 2 a%£#. 

(cos. x) 5 (cos. x) 3 

Ans. — z — — 5 — . 

5 3 

Ex. 2. Integrate the expression (sin. #) 5 (cos. xydx. 

(cos. x) 5 2(cos. x) 1 (cos. a?) 9 
Ans. - g + jj - g . 

Ex. 3. Integrate the expression (sin. #) 4 (cos. xfdx. 

(sin. xf (sin. x) 1 
Ans. — k— — - — =-*-. 

Ex. 4. Integrate the expression (sin. #) 6 (cos. a?) 3 &?. 

(sin. a?) 7 (sin. x) 9 



Ans. 



9 



213. It is required to integrate the expression (tang. x) m dx. 
Assume tang. x~z. 

Then dx — 



z m dz 



and f(tBxig.x) m dx= I t 



+ z 2 ' 

which is a rational fraction, and may be integrated by the 

method of Art. 173. 

Ex. 1. Integrate the expression (tang, xydx. 

z*dz n 7 - dz 
(tang, xydx = -r- — = = z 2 dz —dz + 7- — - 9 . 
x fe J l + z 2 1 + z 2 

ka 1 (tang, x) 3 
Hence y (tang, xydx— ^ —tang. x+x. 

Ex. 2. Integrate the expression (tang. xfdx. 

(tang. a?) 4 (tang, x) 2 _ 
.Ans. f — —— X — -+log. sec. #. 



214. It is required to integrate the expression x n sin. xdx. 
This expression may be integrated by the method of Art. 187. 
A ssume u — x 1 \ and dv S sin. a?<fe ; 

which give du = nx n ~ l dx, and v=— cos. x. 

Then /as" sin. #cfo = — x n cos. a? + 7ifx n ~ 1 cos. a*fe. (1) 



J 



INTEGRATION OF TRIGONOMETRICAL DIFFERENTIALS. 241 

To integrate the second term in this equation, assume 
u = x n ~\ and dv = cos.xdx; 
which give du = (?i — l)x n ~ 2 dx, and v = sin. x. 
Then fx n ~ l cos. xdx = x n ~~ 1 sin. x— (?i — l)fx n ~ 2 sin. xdx. (2) 

The second term of this expression may be integrated by 
formula (1), and we shall obtain the development 
fx n sin. xdx— —x n cos.x + 7iaf x ~ l sin. x + n(n — l)x n ~ 2 cos. x 
—?i(n — l)(n — 2)x n ~ 3 sin. x— etc. (two plus 
and two minus terms alternately). 
Ex.1. Integrate the expression x sin. xdx. 

Arts, —x cos. x+sin.x. 
Ex. 2. Integrate the expression x 2 sin. xdx. 

Ans. —x 2 cos. x + 2x sin. x + 2 cos. a?. 
Ex. 3. Integrate the expression x 3 sin. xdx. 

Ans. — x 3 cos. x + 3x 2 sin. a? + 6a? cos. x — 6 sin. a?. 
Ex. 4. Integrate the expression x* sin. xdx. 
A?is. -x^ cos.x+ix 3 &in.x-\-12x 2 cos.x-24:X ^m.x-24: cos.x. 

215. It is required to integrate the expression x n cos. xdx. 
Proceeding as in the last Article, we shall find 

fx n cos. xdx=x n sin. x + nx n ~ l cos. x— n{n— l)x n ~ 2 sin. x 
— n(?i — l)(n—2)x n ~ 3 cos. x+ etc. (two plus 
and two minus terms alternately). 

Ex. 1. Integrate the expression x cos. xdx. 

Ans. x sin. a?+cos. x. 
Ex. 2. Integrate the expression x 2 cos. #&?. 

-4?is. a? 2 sin. x+2x cos. #— 2 sin. x. 
Ex. 3. Integrate the expression x 3 cos. a%fe. 

Ans. x 3 sin. x + 3x 2 cos. a?— 6a? sin. a?— 6 cos. x. 

216. Inverse trigonometrical differentials. Differential ex- 
pressions containing a?, dx, and sin. -1 a?, or cos.^a?, or tang. -1 a?, 
etc., may frequently be integrated by methods similar to those 
already explained. 

L 



242 INTEGRAL CALCULUS. 

Ex.1. Integrate the expression x 2 sin. -1 ^^. 
Integrating by parts (Art. 187), place 

x 2 dx=dv, and sin.~ 1 x=:u. 

Then v =fx 2 dx = -5-, and du = , . 

Substituting in the formula 

fvdv == w# —fvdu, 

we have /a? 2 sin.'" 1 a?rfa3=-^- sin. _1 #— / — . 

Integrating by Formula A, Art. 192, we have 

C x 3 dx (x 2 2\ , 

^VI3P--U+3JVT^. 

^3 1/x 2 2\ 

Hence y# 2 sin." 1 ^^^-^- sin. -1 #+o(-o +«) Vl — # 2 +C. 

Ex. 2. Integrate the expression a? 4 tang r l xdx. 
Place x*dx~dV) and tangr^a^w. 

Then 0=-F" 5 and du~^-, — r. 

5' l + # 2 

Hence 

> 4 tang.-'aafc= -g- tang." 1 *?- J 5(1+3,2) 

=| tang-^-g J ~ | + 2 log. (1+* 2 ) } +C 

Ex. 3. Integrate the expression x tang r^xdx. 

1 + x 2 . x _. 

-4.^5. — 5 — tang. _1 #— o + C. 



SUCCESSIVE INTEGRATION. 243 



CHAPTER V. 

SUCCESSIVE INTEGRATION. INTEGRATION OF FUNCTIONS OF TWO 
OR MORE VARIABLES. DIFFERENTIAL EQUATIONS. 

217. Successive integration. In the examples of the pre- 
ceding Chapters the first differential coefficient has been given 
to determine the primitive function from which it was derived. 
When, however, instead of the first differential coefficient, the 
second differential coefficient is given, then, by a first integra- 
tion we shall arrive at the first differential coefficient, and it is 
only after a second integration that we arrive at the original 
function. If the third differential coefficient were given, it 
would require three successive integrations to arrive at the 
original function, and so on. Since each integration intro- 
duces a constant, it follows that the complete primitive ought 
to contain as many arbitrary constants as it has required inte- 
grations to obtain it. 

A double integration is sometimes indicated by the symbol 
ff, and sometimes by the symbol/ 2 . So also a triple integral 
may be indicated by//, or by/ 3 , and so on. 

Ex.1. Integrate the expression d 2 u = ax 2 dx 2 (see Art. 54). 

This expression may be written 
d 2 u 



dx 



-~ax 2 dx. 



Hence / -T——fax 2 dx\ 

du ax 3 _. 

(xx dx 
Hence du = — o — + Cdx. 



244 INTEGRAL CALCULUS. 

Integrating again, we obtain 



u =i2+ Cx + G ; 

CIX* 

that is, u —ffax 2 dx 2 = y^- + Cx + Co 

Ex. 2. Integrate the expression d 3 u—axdx 3 . 
This expression may be written 
d 3 u ^(d 2 u\ 

.^ d 2 u ax 2 _, 

Hence ^=-2- + C ' 

and finally, as in the former example, 

u=/ 3 axdx 3 =— + —+C , x+C / \ 

Ex.3. Integrate the expression dhi — 14:4:dx*. 
Proceeding as in the last example, we find 

g=72a»+Q B +C'; 

du Cx 2 ' 

^=24^+ — + C^+C-; 

Ct" 3 fl'r 2 
w =/*144daj* - 6a; 4 + -g- + -^- + C"x + 0'". 

1 

Ex.4. Integrate the expression dhc=ax*dx z . 

. Sax^ Cx 2 __ ^ 

iyw. w = -j^- + -g- + Cx + C". 

Ex.5. Integrate the expression d*u=—x~ A dx Ac . 

loff.a? C# 3 CV -... ' _ 

218. Integration of differential functions of two variables. 
If ?j is a function of a? and y where each variable is supposed 
to be entirely independent of the other, we have found, Art. 66, 



DIFFERENTIAL FUNCTIONS OF TWO VARIABLES. 245 

_ du . d u _ 

where Dw denotes the total differential of the function, while 
the first die in the second member of the equation denotes the 
partial differential taken with respect to #, and the second du 
denotes the partial differential taken with respect to y. 
If, then, we have an expression of the form 

du=N.dx+'Ndy, (1) 

where M and N are functions of x and ?/, if du be the total 

differential of the function, we must have 

du _, , du ^ 
-7- = M, and -y-=K 
ax ? ay 

Differentiating the first expression with respect to y, and the 
second with respect to x y we have 

dM. d 2 u , ^N d 2 u 
and 



dy dxdy* dx dydx 

By a method similar to that employed in Art. 62, it may be 
proved that #^ #2 U 

dxdy~ dydx* 

dM dX 
and hence y=^ ( 2 ) 

or, in order that Mdx+Ndy shall be a total differential, it is 
necessary that 

dM d^ 

dy ~~~ dx' 

which is hence called the criterion of integr ability. 

When an expression of the form (1) is to be integrated, we 
first examine whether the criterion (2) is satisfied. If this 
criterion is satisfied, the equation is integrable. To perform 
the integration, we integrate one of the partial differentials 
(as M.dx) with reference to its corresponding variable, as though 
the other variable were constant, and add such a function of y 

as will satisfy the equation -r-=N. 



246 



INTEGRAL CALCULUS. 



Integrating on the supposition that y is constant, we have 

u=/M.dx+C, 

where the correction C is a function of y alone. 

Ex. 1. Integrate the expression du=ydx+xdy. 

Here ^L=y> and N=#; 

dM. dlS 

dy ~~ ~~ dx ' 

Hence criterion (2) is satisfied. 

u =/Mdx — xy + 0. 

Ex.2. Integrate the expression du=2xy 3 dx+3x 2 y 2 dy. 

Here M = 2xy 3 , and N = 3x 2 y 2 ; 

dM n d^ ' 

dy J dx 

Criterion (2) is satisfied. 

u=/Mdx=x 2 y 3 +C. 

Ex. 3. Integrate the expression 

du = {ay 2 — by 3 )dx + (2axy — 3by 2 x)dy. 

Here M = ay 2 — by 3 , and N = 2axy — 3by 2 x ; 

dM oz 9 ^N 
-7- = 2ay — 3by 2 = -j— . 

Criterion (2) is satisfied. 

u —f{ay 2 — by 3 )dx — ay 2 x — by 3 x + C. 
Ex. 4. Integrate the expression 

du = (2y 2 x + 9x 2 y + 8x 3 )dx + (2x 2 y + 3x 3 )dy. 

Ans. u=y 2 x 2 + 3x 3 y+2x* + C 
Ex. 5. Integrate the expression 

du = 2axydx + a# 2 6?y — j/ 3 <fe — 3xy 2 dy. 

Ans. u=ax 2 y—y 3 x+C 



219. Differential equations. A differential equation is an 
equation involving differential coefficients with or without the 
primitive variables from which those differential coefficients 
were derived. Thus in Arts. 68-70 we have found such dif- 
ferential equations as the following : 



DUTEKENTIAL EQUATIONS. 217 

dy—mdx\ 

ydy=2adx; 

2ydy = vidx + 2nxdx ; 

dy 2 +yd 2 y~ ndx 2 . 

220. Classification of differential equations. Differential 
equations are classified according to their order and degree. 

The order of a differential equation is the same as the order 
of the highest differential coefficient which it contains. 

The degree of a differential equation is the same as the de- 
gree to which the differential coefficient which marks its order 
is raised, that coefficient being supposed to enter into the equa- 
tion in a rational form. 

Thus the equation -/ = — is an equation of the first order 
1 dx y u 

and the first degree. 

The equation l-^J + a(-j-)=b is of the first order and of 

the second degree. 

d 2 u I uij\ 
The equation -t^ -\-a[-j-) + by 2 =c is of the second order 

and of the first degree. 

(d 2 y\ 2 (d 2 y\ 
The equation (y^j + ci\-j-^)—b is of the second order 

and second degree. 

221. Differential equations of the first order and degree 
betiveen tioo variables. 

Differential equations of the first order and degree between 
two variables may be represented under the general form 

™ ^dy A 

M and N being functions of the variables x and y. 
This equation is commonly written 

Mdx+my=0. (1) 



248 INTEGRAL CALCULUS. 

There are certain cases in which equation (1) admits of 
solution. 

222. Case I. When the variables in M and N" can be sepa- 
rated. Equation (1) can be solved when the variables in M 
and N" admit of being separated ; that is, when the equation 
can be reduced to the form 

Xdx+Ydy=0, (2) 

in which X is a function of x alone, and Y a function of y 
alone. 

To solve the equation in its reduced form (2), it is only 
necessary to integrate the two terms separately, and to equate 
the result to an arbitrary constant. Thus the solution will be 
fXdm+/Ydy=C. 

Ex.1. Solve the equation xdx + ydy=0. 
By integration, we obtain 

2 i ~2 ' 
or, since is arbitrary, we may write it 

x 2 +y 2 = C. 

Ex. 2. Solve the equation ydx—xdy—0. 
Dividing by xy, we have 

dx dy 

x - y % 

Integrating, we have log. x = log. y + C ; 
or we may write it 

log. x — log. y + log. a = log. ay. 

Hence x — ay. 

i 
Ex. 3. Solve the equation {l+x^dy—y^dx. 

dy dx 
Here 

Integrating, we have 



y* 



~l+x 2 



%y^ =tang. _, a?+C. 



DIFFERENTIAL EQUATIONS. 249 

Ex.4. Solve the equation xy 2 dx+dy—Q. 

nil 
Dividing by y 2 , xdx + -j[ = 0. 

y 

X 2 1 

Integrating, we have — — - = C ; 

*/ 

or x 2 y— 2 = 2Cy. 

uX (71J 

Ex. 5. Solve the eouation T - l — +q— — =0. 

Ans. (l+x)(l+y)=G. 

223. Case II. When one side of the equation may be ren- 
dered an exact differential, and the other a function which 
contains only one of the variables. 

t* '-• « i xi .• xdy + ydx T 

JjiX.l. bolve the equation — - — - — =—dy. 

x 

Dividing by y, we have 

xdy+ydx dy 

~xy~ ~~~"y' 
Integrating, we have ^ 

log.(a^)=log.-+C. 

Hence %y=-<> 

if 

or xy 2 =a. 

Ex.2. Solve the equation ydx=y 3 dy+y 2 dy+xdy. 
Transposing, we obtain 

ydx—xdy 

t =ydy+d y . 

X 7J 2 

Integrating, we have -=—+?/+ C. 

is 

Ex. 3. Solve the equation ydy—xdx=a{xdy+ydx). 
Multiplying by 2, we have 

d(y 2 — x 2 ) = 2ad(xy). 
Hence y 2 —x 2 — 2axy + C. 

L2 * 



250 INTEGRAL CALCULUS. 

Ex. 4. Solve the equation adx 2 =ydy 2 —bdxdy. 
This equation may be reduced by completing the square, as 
in an algebraic equation of the second degree, and we have 
7 , b 7 _ b 2 dy 2 4ay+&, . 

_ My (Uy+b 2 )* 1 
Hence <*>+■&= 2a * »• 

Therefore x+^J^^+G. 

2a 12a 2 

224. Case III. When the equation is homogeneous. The 
differential equation Mdx+~Ndy=0 is said to be homogeneous 
when M and N are homogeneous functions of x and y, and are 
of the same degree. To integrate a homogeneous equation, it 
suffices to assume x—vy, or y—vx. In the transformed equa- 
tion the variables x and v will then admit of separation. 
Ex.1. Solve the equation y 2 dy=3yxdx—x 2 dy. 
Substitute vy for x, and we have 

y 2 dy — 3y 2 v(vdy + ydv) — y 2 v 2 dy. 
Dividing by y 2 , 

dy = 3v(ydv + vdy) — v^dy = 3yvdvJ- 2v 2 dy. 
dy 3vdv 



Hence 



y 1 — 2v 2 ' 
e 



Therefore log. y=| log. j— -^ ; 

c 3 c 3 y 6 

v = (l-2v 2 ) 3 = (y 2 -2x 2 f 

2 

or y 2 —2x 2 — cy s . 

Ex.2. Solve the equation y 2 dx + xydy+x 2 dy=0. 

Put x—vy. 

Then y 2 ^y + ^ 3 dfe + y Wy + y 2 v 2 dy = 0, 

or 2vdy+v 2 dy+ydv = 0, 

dy dv dv dv 



or 



2/ "" 2^ + ^ 2 ~2(2 + ^) 2*;' 






DIFFEKENTIAL EQUATIONS. 251 

Hence Q /2T~" 

log. y=| log. (2+^-4 log. v=$ log. -^=log. V "1T+ - 



Therefore 


# y=K „ J ; 






2+*> 2*,+* 

y V X 


.a 2 ; 




y 2 x=(2y+x)a 2 . 




Ex. 3. Solve the equation x 2 dy—y 2 dx-. 


-xydx. 


Put 


y—vx. 




Then 


x 2 vdx + aj 3 ^; — v 2 x 2 dx = 


x 2 vdx. 


Hence 


xdv=v 2 dx, 




or 


dx dv 

X ~~ V 2 ' 





1 X 

Hence log. #= — - + <?= — - + C. 

& v y 

225. Geometrical applications of differential equations. 
By means of the preceding principles we are often able to 
deduce the equation of a curve which possesses some remark- 
able property. 

Ex.1. Determine the curve who c 3 subnormal is constant. 

udv 

The general expression for the subnormal (Art. 93) is ^y^. 

Denoting the constant by a> we have the equation 

dx 
Whence ydy = adx ; 

y 2 

and integrating, 77 = ax + 0, 

or y 2 = 2ax+2C, 

which is the equation of a parabola. 

Ex. 2. Determine the curve whose subtangent is constant. 

i/dtjc 

The general expression for the subtangent (Art. 91) is -j-. 



252 INTEGKAL CALCULUS. 

dy 



TT ydx 

Hence ^r- = a, 



ady 
or dx— — -. 

y 

Therefore x = a . log. y + C, 

which is the equation of the logarithmic curve, the modulus of 

the system being a. 

Ex. 3. Determine the curve whose subtangent is double the 
corresponding abscissa. 

t~t i yd® r, 

Here we have -j——2x. 

dy 

TTT1 2dy dx 

Whence — -=— . 

V x' 

or 2 log. y — log. x + C, 

or y 2 = ax, 

which is the equation of a common parabola. 

Ex. 4. Determine the curve whose subtangent is three times 

the corresponding abscissa. 

ydx rt 

dy ' 

Sdy dx 

y ~~ x ' 

3\og.y=log.x+C, 

or y 3 =ax. 

Ex. 5. Determine the curve whose subtangent is n times the 

corresponding abscissa. Ans. y n ~ax. 

This Chapter exhibits a few of the simplest properties of 
differential equations. The student who desires to pursue the 
subject further is referred to larger treatises, such as those of 
Young, Todhunter, Jephson, De Morgan, Price, Boole, and 
others. 



QUADRATURE OF PLANE SURFACES. 253 



CHAPTER VI. 

QUADRATURE OF PLANE SURFACES. RECTIFICATION OF PLANE 
CURVES. SURFACES AND VOLUMES OF SOLIDS OF REVOLUTION. 

Quadrature of Plane Surfaces. 

226. The quadrature of a surface is finding the measure of 
its area, or finding a square equivalent to a proposed area. 
Thus the quadrature of the circle consists in finding a square 
which contains the same area. When the area limited by a 
curve can be expressed in a finite number of algebraic terms, 
the surface is said to be quadrable. 

We have found (Art. 122) that the differential of any plane 
area referred to rectangular co-ordinates is 

dA=ydx, (1) 

where A represents the area included between the curve and 
the axis of X, and x and y are the general co-ordinates of the 
curve. The differential of the area included between the 
curve and the axis of Y is 

dA=xdy. (2) 

If either of these formulas can be integrated, we shall have an 
expression for the value of A, which is the required area. 

Hence, to determine the area limited by a curve and either 
of the co-ordinate axes, we differentiate its equation, and de- 
duce the value of dx or dy, which we substitute in equation 
(1) or (2). Having obtained an expression for the differential 
of the area in terms of one variable and its differential, we 
integrate it, and obtain a general expression for an indefinite 
portion of the area. To obtain the area of a definite portion, 
we must take the integral between the limits denoted by those 
values of the variables which correspond to the extremities of 
the limiting curve. 



254 



INTEGRAL CALCULUS. 



Ex. 1. Find the area of the common parabola. 
Differentiating the equation ^ 2 =4aa?,we have 

ydy 



dx=' 



Whence 



and 



dA-. 



2a' 
y 2 dy 



2a 



'6a 



+ 0. 



(1) 







Is 


x^- 


] 


5^^ 






v 


R 


I 


^ 


v^ 


, 











If we estimate the area from the origin 
O, when y—0^ A=0, and consequently 
C=0, and the corrected integral is 

,3 



A= 



y 

6a' 



which equals 

y 2 x y 4taxy 2xy 
6a = ~6a~ =z ~~3~> 
that is, the area OPR is -§ ODPR ; consequently 

POP'^fDPP'D', 
or the area of the parabola is equal to two thirds of the cir- 
cumscribed rectangle. 

If it is required to find the area intercepted between two 
known ordinates, as PR and MN, that is, the area of PMNR, 
put PR — b and MK=c; substitute b and c successively for y 
in equation (1), and subtract the former result from the latter ; 
that is, integrate between the limits y—b and y=c, and we 
have o?— b 3 



PMNR: 



6a 



Ex. 2. Find the area of a circle. 

The equation of the circle is 

y=(r 2 — x 2 ) 1 *. 
Expanding the second member of this equation by the bino- 
mial theorem, we have 



y=r- 



ar 



ar 



3.x 6 



3.5z 8 



2r 2Ar 3 2.4. Or 5 2.4. 6. Sr 1 ' 



etc. 



QUADRATUKE OF PLANE SURFACES. 



255 



Multiplying this by dx y and integrating each term separately, 
we obtain 



x° 



Sx 1 



3.5z 9 



-TX- 




2.3r 2.4. 5r 3 2.4. 6. lr b 2.4.6.8.9/> 7 
If we estimate the arc from the point 
D,when x=0 9 the area is 0, and conse- 
quently C = 0. The preceding series 
therefore expresses the area of the seg- 
ment CDEH when x equals OH. 

H the arc DE be taken equal to 30°, 

T 

CH or x becomes ~, and we have 

2 jl_ _J_ _ 1 _ 3 

' r \2 2.3.2 3 2.4.5.2 5 2.4.0.7.2 7 " 
The sum of this series may be found as follows : 

2^=0-0208333; 

2X5^ =0 - 0007812; 

2.4.6.7.2 7 =0 - 0000698; 
= 0.0000085: 



CDEH: 



2. 4. 6. 8. 9. 2 9 
3.5.7 
2.4. 6. 8. 10. 11. 2 U= 
etc., 



: 0.0000012, 
etc. 



+c. 



etc. > 



Sum, =0.0216940. 

Therefore CDEH =^{0.5 -0.021694} = 0.478306r 2 . 

But the area of the triangle CEH=— 1-=0.216506^ 2 . Hence 

the area of the sector CDE = 0.261 800/ 12 , which, multiplied by 
12, gives for the area of the whole circle 

3.1416/* 2 , 
which is a little too great, in consequence of the omitted terms 
of the series, and omitted portions of the decimals. 



256 



INTEGEAL CALCULUS. 



Ex. 3. Find the area of an ellipse. 

The equation to the ellipse referred to its axes is 



* a ' 

and consequently the area of the semi-ellipse will be equal to 

b i 

fydx = -f{a 2 —x 2 ydx. 

i 
But (a 2 —x 2 ) 2 dx is the differential of the area of a circle whose 

radius is a (Art. 122); hence the area of the ellipse =-xthe 

area of the circumscribing circle. But the area of the circum- 
scribing circle is tto 2 ; hence the area of the ellipse is equal to 

wa 2 x -, or nab. 

Ex. 4. Find the area of an hyperbola. 
The equation to an hyperbola referred to its centre and 
axes is z 



y 



= -Vx 2 — 



a 



a 6 



b i 

consequently dK = ydx = -(x 2 — a 2 )~*dx. 

Integrating according to Ex. 5, Art. 197, we obtain 



bx, 



ab 



A=—(x 2 -a 2 Y— g- log. {x+ Vx 2 -a 2 } +G. 

In order to determine the constant 
C, make x = CA=a, in which case 
A=0, and we have 




that is, 

Hence area APR 



= -g- log. a. 



xy ab , x + (x 2 — a 2 )* 



QUADRATURE OF PLANE SURFACES. 257 

Hence the entire segment APP' equals 

or xy-ab log. j —^ j - 

Ex. 5. Find the area of an hyperbola whose base is 24, and 
altitude 10, the transverse axis being 30. Ans. 151.6873. 

Ex. 6. Find the area of a cycloid. 

The area of the figure ABC is most conveniently obtained 
by first finding the area of the figure t> e b 

ABD, contained between the lines AD, 
DB, and the convex side of the curve. 

Let BC = 2?>, AG=x, and FGc=y, 
whence FE = 2^— y. We shall then 
have 

dA=d(ADEF) = (2r-y)dx. A ** 

But the differential equation of the cycloid (Art. 144) is 

ydy 



dx— 



V2ry—y 2 



Hence dA = dyV2ry—y 2 , 

and A =.fdy V2ry ~y 2 + C. 

But this is evidently the area of a segment of a circle whose 
radius is r, and abscissa y (Art. 122) ; that is, the area of the 
segment CHI. If we estimate the area of ADEF from AD, 
and the area of the segment CHI from the point C, they will 
both be when y=0; the constant C will then in each case 
be 0, and we shall have 

ADEF = CHI; 
and when y=2r, 

7TT 2 

ADB=the semicircle CHB=-^-. 

But the area of the rectangle ADBC is equal to 
AC x AD = wr x 2r = 2-kt 2 . 

O 2 

Hence the area AFBC = ADBC— ADB = -^-= three times 
the semicircle CHB; hence the area included between one 



258 



INTEGKAL CALCULUS. 



branch of the cycloid and its lase is equal to three times the 
area of the generating circle. 

Ex. 7. Find the area of the logarithmic curve whose equa- 
tion is x=log. y. 




Here dx-. 



M.dy 



dA = ydx = Mdy. 
Hence A=My+G. 
If we estimate the area from 
the line AB, for which x = 0, 
and y=lj we have 

0=M+C; hence C = —M. 
Hence the corrected integral is 
A=My-M=M(y-l). 
If y = 0, we have area ABD = — M. 

If y =2, we have area = M. 

Ex. 8. Find the area of the curve whose equation is y=x—x 3 . 

Substituting the value of y in 
the formula Art. 226, we have 
dA = xdx — x 3 dx. 




Hence 






If we estimate the area from the 
origin, we shall have A = when 
x=0, and consequently C=:0. 
If we make x = OD = 1, we shall 

have areaOBD=j. 

To find the area of PDR, we integrate between the limits 

3 
x=OT) = l, and x—OH=^ and find 

areaPDE-?-^ — 2 

Ex. 9. Find the area of the curve whose equation is 
a 2 y 2 — a 2 x 2 + x" = 0. 



QUADRATURE OF POLAR CURVES. 



259 



dA=-(a 2 -xydx: 



If we estimate the area from the origin, 
A=0 when #=0, and 







Hence the corrected integral is 



A a 6 i /a 3 



If we make #=a, then 



A = area OMD = 

2a 2 



a 6 



3' 



Aa 2 



Hence the oval OMDN=:-^-, and the area of both ovals = : « - 
Ex. 10. Find the area of the curve whose equation is 



a y-a 2 b 2 x 2 +b 2 x*=0. 



Ans. ~ab. 



227. Quadrature of polar curves. 

The differential of the area of a polar curve included be- 
tween a radius vector and the initial line (Art. 126) is 

7A r 2 dO 

tfArz-2-. 

Ex. 1. Find the area of the spiral of Archimedes. 

in 

The equation is r=cr (Anal. Geom., Art. 253). 
Hence d6- 



dA= 



b ' 
irr 2 dr 



If we estimate the area from the pole, we have A=0 when 
r=0, and consequently C = 0. The corrected integral is there- 



fore 



A= 



Trr 



3b' 





260 INTEGEAL CALCULUS. 

At the end of one revolution of the 

radius vector, r = AP = b, and A = the area 

irb 2 
of the first spire = AMP=-r-=| the area 

M ' of a circle described with PA as a radius. 

At the end of the second revolution, r=2b, and the area traced 

Q 7 2 

by the radius vector =—o~- But in the second revolution the 

radius vector describes the part PMA a 
second time; hence to obtain the area 
PNB, we must subtract that described 
during the first revolution. 

tj To 

Hence the area PNB=-r-. 

The area of the second spire is 27r£ 2 , 
or twice the area of the circle whose radius is equal to the 
radius vector at the end of the first revolution. 
Ex. 2. Find the area of the hyperbolic spiral. 
The equation of the hyperbolic spiral is rO=a. 

adr 

A ar 

Hence ^-"Ip 

The subtangent of this spiral is — a (Ex. 2, Art. 105). Hence 

(XT* 

the area described by the radius vector from to r is -^-, which 

is equal to the triangle formed by the radius, the tangent, and 

the subtangent. 

The area included between the curve and two radii vectores 

a(r—r') 
r, r is equal to — ^ . 

Ex. 3. Find the area of the logarithmic spiral. 

e=io g .£ 

Mrdr 
dA= : 



QUADRATURE OF POLAR CURVES. 



261 



If we estimate the area from the pole, where f= 0,0 = 0, and 

Mr 2 
the corrected integral is —r— . 

For the Naperian logarithmic spiral, M = l, and hence the area 
of the Naperian logarithmic spiral is equal to one fourth the 
square described upon the radius vector. 

The area included between the curve and two radii vectores, 

M 

b and r, drawn from the pole is x( r<2 "~^ 2 )- 



228. The area included between two curves. Let BCDE, 
BGHE be two curves whose equa- 
tions are respectively 

y—fx and y—Yx. 
The area LCDM, included by the 
curve BCE and the two ordinates CL, 
DM, is given by the formula 






hi 



£ydx, or f a fx.dx, 



where a and b denote the abscissas of the points C and D. 

So also the area LGHM, included by the curve BGE and 
the same ordinates, is given by the formula 

Fa? . dx. 



/ 

Ja 
Hence the area CDHG is given by the expression 

A= I fx.dx— I Fx.dx, or/ (fx—Fx)dx. 
Ja J a Ja 

If we require the area BCEHB, included between the two 
curves BCE and BGE, we draw ordinates through the points 
of intersection B and E, and make a equal to the abscissa AK, 
and b equal to the abscissa AN. 

If the proposed area is bounded by the curves x—fy and 



262 INTEGRAL CAXCULUS. 

x—Yy^ and straight lines parallel to the axis of x at distances 
respectively equal to c and d, we have in a similar manner 

A=ff(f< / -Yy)dy. 

Ex. 1. Find the area included between the straight line 
y=2x and the parabola y 2 =12x. 

The two lines intersect where x=0,y=0, and also where 
x=3,y=6. 2 

The area of the parabolic space (Art. 226, Ex.1) is -~xy =12. 

The area of the triangle is ^xy=9. 

The area of the space included between the proposed lines 
is therefore 3. 

The student should construct a figure illustrating the ex- 
ample. 

Ex. 2. Find the area included between the straight line 
y=z2x— 8 and the parabola y 2 — Sx. 

The lines intersect where x = 2, y~ — 4, and # = 8, y = 8. 

The area of the required space above the axis of abscissas 

• 128 1* 80 
1S — -16=-g. 

The area of the required space below the axis of abscissas 
. 16 28 

IS -w- + 4:=^~. 

™ . . 108 M 

The entire area is -5- = 36. 

Ex. 3. Find the area included between the parabola y 2 = 8x 
and the circle y 2 — lQx— x 2 . 

The curves pass through the origin, and meet at the point 
3=8, y=8. 

If we take only the area which lies on the positive side of 
the axis of x, we have 

/•8 ( / /— ) 7 64tt 2 nA Hn 128 

A= / J Vl§x—x 2 -V$x [dx=-£- -g.64 = 167r--3-. 

256 
The whole area will therefore be 327r— -tt-. 

The student should construct the figure. 



AREA INCLUDED BETWEEN CURVES. 



263 



229. Area included between different branches of the same 
curve. The method of the preceding Article is applicable to find- 
ing the area included between two branches of the same curve. 

Ex.1. Find the area included between the two branches of 
the curve y 2 + x 2 -±x-8y+lQ = Q. 

We find y = 4 db V±x— x 2 
Hence we may put 



Jx — 4:+V4:X—x 2 ; 

Fx=4:— V^x—x 2 . 

fx—Fx=2V4:x—x 2 




Air. 



Then 

and the complete area of the curve is 

J2V4:X— x 2 .dx~ 


Ex. 2. Find the area included between the opposite branches 
of an hyperbola, its transverse axis, and any line parallel to it. 

_ _ ( ay + bx 
Ans. xy+ao log. -j r — 

The preceding method may be extended so as to give the 

area included between anv number of curve lines. If we 

«/ 

have two closed curves intersecting each other, as in the 
figure, and we wish to determine the 
area of BCDE, we must determine the 
points of intersection from the equa- 
tions of the curves, and integrate for 
each portion of the bounding lines of 
the desired area. The area may be 
divided by straight lines parallel to 
the axes into parts the values of which can be separately ob- 
tained by means of the preceding principles. 

230. Area referred to oblique axes. Hitherto we have sup- 
posed the axes to be rectangular ; but if they are oblique, and 
inclined-at an angle w, the formula in Art. 122 becomes 

<#A = sin. toydx. 
Hence A = sin. wfydx. 




264 INTEGRAL CALCULUS. 

For example, the equation to the parabola when the axes are 
a tangent to the curve and a diameter drawn through the point 
of contact, is y 2 = ka f x. 

Hence the area included between the curve, the axis of a?, and 
an ordinate at the point for which x—c^ 

3 

Co n—r 7 4 sin. wJa'c* 
sm. w / y^ax.dx—— 



J' V&a'x.t 




3 ' 

that is, two thirds of the parallelogram which has the abscissa 
c and the ordinate at its extremity for adjacent sides. 

Rectification of Plane Curves. 

231. To rectify a curve is to determine its length, or to 
obtain a right line equal in length to any assigned portion 
of the curve. When the length of an arc of a curve can be 
expressed in a finite number of algebraic terms, the curve is 
said to be rectifable. 

We have found (Art. 121) that the differential of an arc of 
any plane curve referred to rectangular co-ordinates is 

dL=Vdx 2 + dy 2 , (1) 

where L represents the length of the curve, and x and y the 
general co-ordinates. If this formula can be integrated, we 
shall have an expression for the value of L, which is the length 
of the arc. 

Hence, in order to rectify a curve given by its equation, we 
differentiate its equation, and deduce the value of dx or dy, 
which we substitute in equation (1). Having obtained an ex- 
pression for the differential of the arc in terms of one variable 
and its differential, we integrate it, and obtain a general ex- 
pression for an indefinite portion of the arc. To obtain the 
length of a definite portion of the curve, we must take the in- 
tegral between the limits denoted by those values of the vari- 
ables which correspond to the extremities of the given portion 
(Art. 165). 

Ex.1. Rectify the semi-cubical parabola whose equation is 
y 2 = ax 3 (Anal. Geom., Art. 237). 




KECTIFICATION OF PLANE CURVES. 265 

Differentiating the given equation, we obtain 

_ 3ax 2 dx 
dy — — o . 

TTT~I T O *J (ZvL/UL'vL/ 

Whence dy 2 — — j — . 

Substituting this value in equation (l),we have 

7T / , m 9axdx 2 Y f H 9ax\* r 
dL=[dx 2 + 4 J =[l+-^-J dx. 

Integrating by Art. 158, we have 

(4+9aaQ* 
L= 27a +0 - 
To determine the constant C, we may reckon the length of 
the curve from the origin, where L = when x=0, and we have 

g 

"Whence C= 



27a 

Consequently the corrected integral is 

(4:+9axj*— I 



L=- 






27a 

which expresses the length of an arc of the curve from the 
origin to the point whose abscissa is x. 

Ex. 2. Eectify the curve whose equation is y 2 =x> an( ^ deter- 
mine the length of the curve from the origin to the point whose 
abscissa is 10. Arts. 19.0248. 

Ex. 3. Eectify the common parabola whose equation is 

y 2 —4ax. 

Differentiating;, we find dx 1 — t 9 . 
to? 4<z 2 

Whence dL = -£(y 2 + 4a 2 j*. 

Integrating as in Ex. 5, Art. 197, we obtain 

L= yVy l^ a \ a log. (y+ Vtf + W) + C 
M 



266 INTEGRAL CALCULUS. 

If we estimate the arc from the vertex of the parabola, we 

shall have y=® when L = 0. 

Hence = a log. (2a) + C, or C = — a log. (2a), 

and the corrected integral is 

yVy 2 + U 2 ( y+Vy 2 +4:a 2 ) 

L = —^a +a l0 ^ { 2a \ ' 

Ex. 4. Determine the length of the arc of a parabola cut off 
by a double ordinate to the axis whose length is 12, the ab- 
scissa being 2. Arts. 12.8374. 

Note. — Nap. log. =com. log. X 2.3025. 

Ex. 5. Rectify the circle. 

Differentiating the equation x 2 + y 2 =r 2 ,we have 

_ . x 2 dx 2 

Whence dL= — ==. 

Vr 2 —x 2 

Putting r=l, and developing in a series, we have 

7T _ x 2 dx 1.3x*dx 1.3.hx*dx 
dL=dx + -^-+ 2A + 2A 6 + etc. 

Integrating, we obtain 

x 3 1.3^ 5 1.3.5^ 7 

If we measure the arc from the axis of ordinates, we shall 
have L = when a? = 0, and therefore C = 0. 

If we make x = r = l, we shall have for the length of a 
quadrant w l 1.3 1.3.5 

2 = 1 + 273 + 2J~5 + 2.4.6.7 + CtC * 
This series converges so slowly that it requires a very large 
number of terms to furnish an accurate value of tt. 

If in equation (1) we make x = \, then L will correspond to 

IT 

an arc of 30°, that is, to 77, and we shall have 

r 1 1 1.3 •• .• ••«*.*» , _ 0523599 
6~2 + 2.3.2 3 + 2.4.5.2 5 + 2.4.6.7.2 7 + v.omow. 

Hence v= 3.14159. 



KECTIFICATION OF PLANE CUKVES. 267 

232. A better method of obtaining the value of ir is to find 
the length of the arc in terms of its tangent. 

We have found (Art. 53) that if L represents an arc of a 
circle, and x its tangent, we have 

dx 
1 + x 2 
Developing by the binomial formula, we have 
dL=dx(l — x 2 + x i — x 6 + etc.). 
Integrating and determining C as before, we obtain 

/yj3 /-y>5 /yj7 /y>9 

___ tAj Us tO Us /r\\ 

~~3~ t ~5~~~7 r + ~9'~ etc ' ^ 

If we take L equal to an arc of 30°, its tangent will be equal 
to Vt? which, being substituted for x in formula (2), gives 

6 = V 3 x(1 ~0 + O i ~7^ 3 + O i ~-'' ): " a523599 - 

A still more convenient formula for computing the value of 
7r has already been given in Art. 58. 

Ex. 6. Eectify the ellipse. 

For this purpose w-e employ the equation (Anal. Geom., 
Art. 127) y 2 - 'Q-ffat—d). 

Differentiating, we obtain 

(l-e 2 )x 2 ■ 9 
dtf= 2 \ dx 2 . 

Substituting this value in the differential of the arc, we obtain 



7T _ a f (l-e 2 )x 2 dxVa 2 —e 2 x 2 
dL = dxyl+~ 



a 2 —x 2 ' <y/ 



a'—x 4. 



adx 



Va 2 -x* y ' <* 



>/!-?• 



■\A^ 



Developing yl- — j- in a series, we obtain 

adx j e 2 x 2 e*x* Se 6 x 6 



}• 



268 INTEGRAL CALCULUS. 

The several terms of this series may be integrated by 
Formula A, Art. 192, and we obtain 

e 2 e* 3e 6 

where X represents the circular arc whose sine is -. 

aX Q x 

T~ 2 1 



X 2 =-^-„Va 2 -x 2 



X 6 =-T-X 4 — 77V a 2 — x 2 , etc. 

If we estimate the arc of the ellipse from the extremity of 
the conjugate axis, we shall have L = w T hen a? = 0, and conse- 
quently C = 0. 

In order to obtain one fourth of the circumference of the 
ellipse, we must integrate between the limits x — and x = a. 
But when x — a, Va 2 — x 2 = 0; hence the values of the quanti- 
ties X 2 , X 45 etc., become 

X--X • 
Sa* 3a* 

_5^ 3.5a 5 

6 ~ 6 4 ~~2.4.6 °' C,? 
and consequently equation (1) becomes 

i e 2 3e* 3.3.56 6 ) 

°( 2.2 2. 2. 4. 4~2. 2. 4. 4. 6.6" etc ' j ' 
which expresses the length of one fourth of the circumference 
of the ellipse, and X represents one fourth of the circumfer- 
ence of the circle whose radius is a. Hence the entire cir- 
cumference of the ellipse is equal to 

( "> 3e* 3.3.5e 6 ) 

27r ^| 1 "2.2~2.2.4.4""2.2.4.4.6.6~ etC ' \' 



i 



RECTIFICATION OF PLANE CURVES. 269 

This series is always converging, and converges very rapidly 
when the eccentricity is small. 

Ex. 7. Determine the circumference of an ellipse whose two 
axes are 24 and 18 inches. Ans. 66.31056. 

Ex. 8. Rectify the cycloid. 

The differential equation of the cycloid (Art. 144) is 

dx=^=, 

VVry-y 2 

Substituting this value of dx in the differential of the arc, we 
obtain 

Integrating by Art. 158, we obtain 

L=-2(2r)*(2>--y)*+C. 
If we estimate the arc from the point B, where y=2r,we shall 
have = + C, or C = 0, 

and the corrected integral is 

L=-2V2/'(LV-y), 
which expresses the length of the arc of the cycloid from B to 
any point D whose co-ordinates B 

are x and y. But in the right- /^~6 

angled triangle BGC /^ wnm &\ 

BG = VBCxBE= V*r(2r-y). L J* 

Hence arcBD = -2BG, A c E 

or the arc of a cycloid, estimated from the vertex of the axis, 
is equal to twice the corresponding chord of the generating 
circle. Hence the arc BDA is equal to twice the diameter 
BC, and the entire arc ADBH is equal to four times the diam- 
eter of the generating circle. 

233. Rectification of plane curves referred to polar co-or- 
dinates. 

The differential of an arc of a curve referred to polar co- 
ordinates (Art. 125) is 

dL=Vd?* 2 +r 2 dQ 2 . (1) 



270 



INTEGRAL CALCULUS. 



In order to rectify a polar curve, we differentiate its equa- 
tion, and deduce the value of dr or dO, which we substitute in 
equation (1). Having obtained a formula with but one varia- 
ble, we integrate it, and obtain an expression for an indefinite 
portion of the arc. 

Ex. 1. Rectify the logarithmic spiral whose equation is 

T 

= log. - = log. r — log. a. 



a 



Differentiating, we have 



dO-- 



Mdr 



Hence, by substitution, we find 

dL=drVl + M. 2 , 

and L=rVT+W+0. 

If we estimate the arc from the pole, so that L = when ^=0, 

the constant O = 0, and the corrected integral is 

L=rVl + M 2 . 

For the Naperian system of logarithms M = l, and hence 
Jj=r-\/2 ; that is, in the Naperian logarithmic spiral the length 
of an arc estimated from the pole to any point of the curve 
is equal to the diagonal of a square described on the radius 
vector. 

If the radius vector revolves in the negative direction, it 
generates a portion of the spiral which slowly approaches the 
pole, but can not reach the pole until after an infinite number 
of revolutions. In this case the length of an infinite number 
of spires of the curve is finite (see Anal. Geom., p. 193). 

Ex. 2. Rectify the spiral of Archimedes whose equation is 

r=aO. 

dr 

Differentiating, we have 



d9- 



a 



Hence 



and 



dL-. 



(r* + a*]*dr 
a 



1 i 

L^-fi^ + a^dr. 



QUADRATURE OF SURFACES OF REVOLUTION. 271 



The process of integration is the same as that used in rectify- 
ing the common parabola (Ex. 3, Art. 231) ; hence 
rVr* + a? a . ( r+ Vr 2 + d 2 ) 

L = — 2^- + 2 l0 ^- I a \ ' 

where the arc is estimated from the pole. 



QUADRATURE OF SURFACES OF REVOLUTION. 

234. We have found (Art. 123) that the differential of the 
surface of a solid of revolution, when the generating curve re- 
volves about the axis of X, is 



dS = 2jryVda? + dy 2 . (1) 

If the curve revolves about the axis of Y, the equation is 

dS = 27rxVdx 2 + dy 2 . (2) 

In order to determine the surface of a solid of revolution, 
we differentiate the equation of the generating curve, and de- 
duce from it the values of y and dy in terms of x and dx, 
which we substitute in formula (1) or (2). Having obtained 
an expression for the differential of the surface in terms of 
one variable and its differential, we integrate it, and obtain a 
general expression for an indefinite portion of the proposed 
surface. To obtain the area of a definite portion of the sur- 
face, we must take the integral between the limits denoted by 
those values of the variables which correspond to the bounda- 
ries of the proposed surface. 

Ex. 1. Determine the convex surface of a cone. 

If the right-angled triangle ABC be revolved about AB, the 
hypothenuse AC will describe the convex 
surface of a right cone. 

If the origin of co-ordinates be at A, 
the equation of AC is 

y =mx. 
Whence dy = mdx, 

and dS = 2wmxdxVl + m 2 . 

Consequently S = 7r??ix 2 Vl + m 2 + C. 




272 



INTEGRAL CALCULUS. 



If we estimate the area from the vertex, where a?=0, we 
have C = 0, and the corrected integral is 
S = irmx 2 -y/l + m 2 . 
Making #=AB=A,we have the convex surface of the cone 
whose altitude is A, 

S = 7rmA 2 Vl+m 2 . 

Put 5=BC, the radius of the base. Then m=T, and we have 

convex surface of cone = irb Vh 2 + b 2 = 2wb x -~-; 

that is, the convex surface of a cone is equal to the circumfer- 
ence of its base into half the slant height. 

Ex. 2. Determine the convex surface of a cylinder. 

If the rectangle ABDC be revolved about 



D 



B 



Hence 
and 



the side AB, the side CD will describe the 
convex surface of a right cylinder. Let 
CA— b\ then the equation of the line CD 
will be 

y=b, and dy=0. 
dS — 27rbdx, 
S = 27rbx+G. 

If we estimate the area from the origin A, we shall have S=0 
when as = 0, and consequently C = 0. 

If we make x = AB = h J we have the convex surface of the 
cylinder — 2irbh\ that is, the convex surface of a cylinder is 
equal to the circumference of its base into its altitude. 
Ex. 3. Determine the surface of a sphere. 
The equation of the generating circle, referred to the centre 
as the origin, is x 2 + y 2 = r 2 . 

dif- 



Wlience 



Hence 



r 



dS 



= 2iryV< 



dx 2 



z dx* 



and 



= 2-rrrdx ; 





QUADRATURE OF SURFACES OF REVOLUTION. 273 

If we estimate the surface from a great circle perpendicular 
to AB, C will equal ; and if we make x=r, 
we have for the surface of a hemisphere 27rr 2 , 
and for the surface of the sphere kirr 2 ; that 
is, the surface of a sj?here is equal to four 
of its great circles. 

The surface included between the planes 
determined by x =x 1 and x = x 2 

is 2irr(x 2 —x 1 ). 

Hence the area of a zone of a sphere is equal to the product 
of its altitude by the circumference of a great circle. 

Ex. 4. Determine the surface of a 
paraboloid. 

A paraboloid is a solid described by the 
revolution of an arc AC of a parabola 
about its axis AB. A~ 

The equation of the parabola is y 2 =4:ax. 

Whence dx 2 ^-^. 

4a 2 

Hence d8=^(4:a 2 +y 2 )^dy y - 

and S = ^a 2 +y 2 f+C. 

If we estimate the surface from the vertex, we shall have 
S = when y—0^ and consequently 

Whence C = — — s— ; 

and the corrected integral is 

surface=^{(4« 2 +3/ 2 )^-8a 3 }, 

8irO* (/ 3 3 

or —^-{(a+xy-d*}. 

M2 



274 INTEGRAL CALCULUS. 

Ex. 5. Determine the convex surface of a paraboloid whose 
axis is 20, and the diameter of whose base is 60. 

Ans. 3848.451. 

Ex. 6. Determine the surface of a prolate spheroid. 

A prolate spheroid is supposed to be generated by the revo- 
lution of an ellipse about its transverse axis. 

d S = 2jry Vdx 2 + dy 2 = 2irydL. 
But in Ex. 6, Art. 232, we have found 

adx j e 2 x 2 e*x* 3e 6 x 6 
dL= Vd^ 2 1 2^~~2^~2.4.6a 6 ~ ( 
nce Ziraydx ( e 2 x 2 e*x* 3e 6 x 6 ) 

^—tf ( X ~2tf-~2Atf~~2A.§a*~ eta J ' 



v; 



a* 



A 



But -,S==b. 

Va 2 —x 2 

7n A 7 7 ( , e 2 x 2 e*x* 3e 6 x 6 ) 

Hence dS = 2 7 rbdx j 1-^^-^--—-,- etc. | ; 

and integrating each term separately, we obtain 

b_27rto | 1 -2^" 2 "2.4.5^~2.4.6.7^ 6 " ' ' • f + ' 
Integrating between the limits #=:0 and #:=#, we shall obtain 
half the surface of the spheroid, 

j _f_ e* 3e 6 3.56 8 ) 

- 27rab \ 1 ~2. 3~2.4.5~2. 4.6. 7~"2. 4.6.8. 9~~ etC * j ' 

The surface of a spheroid may be expressed by means of 
trigonometrical functions without expansion into series; but 
the formula here given is generally more convenient for com- 
putation. 

Ex. 7. Determine the surface of a prolate spheroid whose 
axes are 50 and 40. Ans. 5882.638. 

Ex. 8. Determine the surface of an oblate spheroid. 

An oblate spheroid is supposed to be generated by the revo- 
lution of an ellipse about its conjugate axis. 

This problem may be solved in a manner similar to Ex. 6, 
and the result will only differ in the signs of the alternate 



QUADRATURE OF SURFACES OF REVOLUTION. 275 

terms of the series. "We shall obtain the entire surface equal to 
. I . e 2 e 4 3e e 3.5e 8 ) 

W I 1 + 0-2X5 + 2T4^7-2.4.6.8.9+ etc " \ ' 

Ex. 9. Determine the surface of the earth, assuming it to be 
an oblate spheroid whose axes are 7900 and 7926 miles. 

Ans. 196,926,600 square miles. 

Ex. 10. Determine the surface generated by the revolution 
of a cycloid about its base. 

We have found in Ex. 8, Art. 232, 

dL=(2r)^(2r-y)~^dy. 

Hence dS = 27r(2r)%(2r - y)~ ^dy = 2ir(2r) *y* (2ry - y 2 )~ ^dy, 
the integral of which will give the value of the surface required. 
But according to Ex. 10, Art. 197, 

Hence S = 27r(2r)M -^(2r- y y — f(2r-yf J +0. 

If we estimate the surface from 
the plane passing through B per- 
pendicular to AC, we shall have 
S = when y—2r^ and conse- 
quently C = 0. 

Taking the area between the limits y=0 and y= 2r, we have 

half the required surface = — 5 — . 

-^ , 6471-r 2 

Hence the entire surface = — ~ — ; 

that is, the surface generated by a cycloid revolved about its 

64 

base is equal to -^ times the area of the generating circle. 

Ex.11. Determine the surface generated by the revolution 

of a cycloid about the tangent at its vertex. 32 

Ans. -^nr 2 . 




276 INTEGRAL CALCULUS. 



Ex. 12. Determine the surface generated by the revolution 



of a cycloid about its axis. 

J Ans. 8irr 2 



CUBATURE OF SOLIDS OF REVOLUTION. 

235. The cubature of a solid is finding the measure of its 
solid contents, or finding a cube equivalent to the proposed 
solid. 

We have found (Art. 124) that the differential of the volume 
of a solid of revolution is 

dV=7ry 2 d%, (1) 

where x and y are the co-ordinates of the generating curve, and 
the axis of X is the axis of revolution. 

To obtain the cubature of any particular solid, we differen- 
tiate the equation of the generating curve, and deduce from it 
the value of dx in terms of y and dy, or the value of y 2 in 
terms of x, which we substitute in formula (1). The integral 
will be an expression for an indefinite portion of the solid. 

Ex.1. Determine the volume of a right cylinder. 

C ,D The equation of CD will be 

y=AC = b. 
Hence d Y = iry 2 dx — wb 2 dx, 



B and Y—Trb 2 x + G. 

Taking the integral between the limits x — and x— AB = A, 
we have V = ir b 2 h ; 

that is, the volume of a right cylinder is equal to the product 
of its base by its altitude. 

Ex. 2. Determine the volume of a right cone. 

As in Ex. 1, Art. 234, y — mx. 
Hen ce d Y — iry 2 dx = irm 2 x 2 dx, 

__ 7T?7l 2 X 3 _. 

and Y= g +C. 

If we estimate the volume from the vertex of the cone,C=0. 
Let x—h. 

_. . . 7rm 2 h 3 irb 2 h 

Then the entire cone = — - — = — ^— ; 

o o 

that is, the volume of a right cone is equal to the area of its 






CUBATURE OF SOLIDS OF REVOLUTION. 277 

base into one third of its altitude — that is, is equal to one 
third of the circumscribing cylinder. 

Ex. 3. Determine the volume of a prolate spheroid. 

The equation of an ellipse is 
b 2 

7rb 2 
Whence dY = —(a 2 —x 2 )dx, 



and V='—[a 2 x— ■= j + C 




If we estimate the volume from the plane passing through 
the centre perpendicular to the transverse axis, we shall have 
V=rO when #=:0, and consequently = 0, and the corrected 
integral is irb' 



-(^-i> 



/2 



a 4 

Making x=a, we obtain for one half the volume 

27rb 2 a 

and for the entire spheroid ^ 

iirb 2 a 2tt£ 2 ft 
__ = _ X 2a; 

that is, the volume of a prolate spheroid is equal to two thirds 
of the circumscribing cylinder. 

The volume of a segment of a prolate spheroid included 
between two planes perpendicular to the transverse axis, one 
of which passes through the centre, is 

irb 2 x( x 2 \ 

Ex. 4. Determine the volume of an oblate spheroid. 

7T<2 2 

dY=Ttx i dy=j^{b 2 -y 2 )dy. 

Hence y^^y-fj + C, 

which, between the limits y=—h and ?/= +b, gives 

47r« 2 & 2ttci 2 , 
— = — X25; 

that is, two thirds of the circumscribed cylinder. 



278 



INTEGRAL CALCULUS. 



4nrb 2 a 4ma 2 b 




Hence 
the prolate spheroid : the oblate spheroid : : —^- ; Hl^LZ :: b :a . 
If in either expression we make a = b,we obtain 

4:7TT 3 

the volume of a sphere == — ~— . 

Ex. 5. Determine the volume of a paraboloid. 
d C dY=4nraxdx. 

Hence V = 27rax 2 + C. 

If we estimate the volume from A, 
C=0, and we have 

the volume = 2irax 2 = \-Ky 2 x ; 
B that is, half the area of the base x per- 
pendicular height ; or half the volume of the circumscribed 
cylinder. 

Ex. 6. Determine the volume of a paraboloid whose axis is 
30, and the diameter of its base 40. Am. 18849.556. 

Ex. 7. Determine the volume of a parabolic spindle generated 
by the revolution of a parabola about its double ordinate BO. 
A Let 

AD=A,BD=6, AN=a 7 andPN=y. 
Then PM = A— a?, and the area of the 
circle described by the point P in its 
revolution = ir{h — x) 2 . 
Hence d V = ir{h —x) 2 dy — 7ch 2 dy — 2-rrhxdy + irx 2 dy. 

Substituting for x its value ^-, we have 

7rhy 2 dy mfdy 
dV = irh 2 dy- ^ J + - 




Hence 



V = irh 2 y- 



2a 
irhy 3 wy 5 



16a 2 



6a 



or 



80a 2 

7T 



fO, 



V = irh 2 y — TJthxy + ^yx 2 + C, 



Taking the integral between the limits x = and x=h,we have 

2 To7 7T 7o7 87rA 2 £ 

V = ttA 2 b - T,irh 2 b + ^h 2 b = — r=- ; 



CUBATURE OF SOLIDS OF REVOLUTION. 



279 



that is, the volume of the spindle is eight fifteenths of the cir- 
cumscribed cylinder. 

Ex. 8. Determine the volume of a parabolic spindle whose 
length is 80, and whose greatest diameter is 32. 

Ans. 34314.569. 

Ex. 9. Determine the volume of the solid generated by an 
arc of a parabola revolving about the tangent 
at its vertex. E . *f_ 

The area of the circle described by the point 
P in its revolution about AE is wx 2 . Hence h 



iryhly 
d\ =wx 2 dy = ifi 2 • 



V: 



7rjr 



7rX 2 y 



a 



\A 



N 



D 



Hence 

bOa 2 5 

Taking the integral between the limits x = and x=h,we have 
V = — ^— , or one fifth of the circumscribed cylinder. 

Ex. 10. Determine the volume of the solid generated by an 
arc of a parabola revolving about EB parallel to the axis AD. 

The area of the circle described by the point P in its revo- 
lution about EB is Tr(b—y) 2 . Hence 
dV=7r(b-y) 2 dx 

= 7r b 2 dx — 2-n-bydx + iry 2 dx 

i i 
= wb 2 dx — 4nrba?x~*dx + kiraxdx. 



Hence 



V=7rb 2 x^^rrba 2 x 2 +27rax 2 + G 



TV 



=wb 2 x— T^bxy + -^xy 2 + C. 
Taking the integral between the limits x—Q and x=h,we have 



Y=7rb 2 h^7rb 2 h+^b 2 h, 



rb 2 h 



or V = — £-, or one sixth of the cylinder having the same base 
and altitude. 



280 INTEGRAL CALCULUS. 

Ex. 11. Determine the volume of the solid generated by the 
revolution of a cycloid around its base. 

The differential equation of the cycloid (Art. 144) is 

V^ry—y 1 
Hence dY=7ry 2 dx=-^ 



V2ry—y 2 

This expression may be integrated by Formula E, Art. 196, and 
we obtain 

7TV 2 ! §7TV _ 1 

The integral of y 2 (2ry—y 2 ) 2 dy has been given in Ex. 9, 
Art. 197. 

The integral must be taken between the limits y—0 and 
y=2r. When y = 0, all the terms of the above expression 
become 0. When y = 2r, the arc of which y is the versed sine 
becomes nr, and the expression for Y reduces to 

3r 5-n-r 

which equals — ~ — ; 

that is, the volume of the solid is five eighths of the circum- 
scribing cylinder. 



APPLICATIONS OF THE CALCULUS TO MECHANICS. 281 



CHAPTER VII. 

APPLICATIONS OF THE DIFFERENTIAL AND INTEGRAL CALCULUS 

TO MECHANICS. 

Centre of Gravity, 

236. Definitions. In determining the centre of gravity of 
a body, it is convenient to regard the body as composed of an 
indefinitely large number of particles, each of which is indef- 
initely small, and the weight of the body is the resultant of 
the weights of the different elementary portions of the body 
acting in parallel and vertical lines. This resultant passes 
through a point which is called the centre of gravity of the 
body. 

It is proved in treatises on Mechanics that if each particle 
of a body be multiplied by its distance from any assumed 
plane, and the sum of the products be divided by the sum of 
the particles, the quotient will denote the distance of the com- 
mon centre of gravity from the assumed plane. The product 
of any particle by its distance from a given plane is called its 
moment with respect to that plane. The moment of a body 
is the sum of the moments of its particles. 

237. In all of the following examples bodies will be sup- 
posed to be of uniform density. Surfaces, lines, and points 
are supposed to be material. A material surface is a body 
whose length and breadth are finite, but whose thickness is in- 
definitely small ; a material line is a body whose length is 
finite, but its breadth and thickness are indefinitely small ; and 
a material point is a body whose length, breadth, and thickness 
are indefinitely small. 

238. To find the centre of gravity of a line. 

Let BAC be any plane curve symmetrical with respect to 



2S2 



INTEGRAL CALCULUS. 




the axis AX, and let AY be drawn perpen- 
dicular to AX. Draw PRP' perpendicular 
to AX. Then, since the curve is symmet- 
rical with respect to AX, PR is equal to 
PR, and the centre of gravity of the two 
particles P and P' will be on the line AX. 
The same will be true of any two corre- 
sponding particles on the two branches of 
the curve. Hence the centre of gravity of the line BAC must 
be situated in the line AX. Designate the position of the 
centre of gravity by G ; it is required to find the distance of 
G from AY. 

Let x represent the abscissa, and y the ordinate, for any par- 
ticle of the line BAC. It is convenient to regard the differ- 
ential of the line as representing one of the elementary portions 
or particles of the line. If, then, s represents the line BAC, 
ds will represent one of the elementary particles of the line. 
If we multiply this differential by the distance of the particle 
from AY, the product xds will represent the moment of that 
particle, and its integral foods will represent the sum of the 
products arising from multiplying each particle by its distance 
from AY. Dividing this sum by the sum of the particles, or s, 

we have — — -, which represents AG, the distance of the centre 

of gravity from AY. 

If we consider simply the portion of the line comprehended 
between A and B, the distance of its centre of gravity from 
AY will be the same as that of the whole line BAC, but its 

distance from AX will be represented by — — . 

Ex. 1. Determine the centre of gravity of a straight line. 
Conceive the line to be placed on AX, with one extremity 
at the origin A. The moment of any particle of the line is 
. and the moment of the whole line is J\vclv, while the 
length of the line is X. Hence 

fxdx \&+C x 
"~ x ~2 ; 



AG = 



CENTRE OF GRAVITY. 



283 



that is, the centre of gravity of the line is at its middle 
point. 

In the above expression the constant C is 0, because when 
x=0 the function = 0. The same remark ap- 
plies to most of the subsequent examples. 

Ex. 2. Determine the centre of gravity of a 
circular arc BAC = 2s. A 

The equation of the circle referred to the 
axes AX, AY, is 

y 2 =2rx—x 2 . 

(r—x)dx 




Hence 

and 

Hence AG: 

Hence 



dy- 



V2rx—x 2 ' 



ds = Vdx 2 + dy 2 = 



rdx 



V2rx—x 2 



fxds 



r r xdx 

sJ V%rx — 



rx—x c 



T y 



= -\ — V2rx—x 2 +s\ =r — 



OG = r -AG= 



T y 



that is, the distance of the centre of gravity from the centre of 
the circle is a fourth proportional to the arc, the radius, and 
the chord of the arc. 

When the arc is a semi-circumference, the chord = 2r, and 
the arc = 7r?\ Therefore o 

OG=— . 

7T 

Ex. 3. Determine the centre of gravity of the arc of a 
cycloid. 

We will assume as the axes of reference the axis of the 
cycloid and a tangent to the curve at its vertex. 

The differential equation of the cycloid is 



y . y v y 



284 



INTEGKAL CALCULUS. 



But since the origin is at the vertex, substituting 2r—y for y, 



we have 



dx 2 



-!/ 



y 



■dy 2 



Therefore 

Also, 

Hence 



, /2r 

ds = V dx 2 + dy 2 = dyy — . 



AG=^ 



s=2 V2ry (Ex. 8, Art. 232). 

For the entire curve y=2r, so that the distance of the centre 
of gravity of the entire curve from the vertex is equal to one 
third of the diameter of the generating circle. 

239. To find the centre of gravity of a plane surface. 

Let BAC be a plane curve symmetrical with 
respect to AX ; it is required to find the centre 
of gravity of the surface bounded by this curve 
and the line BC drawn perpendicular to AX. 
The distance of the centre of gravity from AY 
must be the same for the whole figure BAC as 
for the half figure BAD. 

The differential of the area of the segment 
is ydx. Regarding the surface as a material body, and multi- 
plying the differential of the area by x, the distance of any 
particle from AY, the product xydx will be the differential of 
its moment. Hence the moment of the entire surface BAD' 
will be represented by /xydx, and the distance of the centre of 
gravity from AY will be rej)resented by 

fxydx 
area 
If we consider simply the area BAD, the distance of its 
centre of gravity from AY will be the same as that of BAC ; 
but to find the distance of the centre of gravity of BAD from 
AX, we must multiply ydx by the distance of the centre of 
gravity of the element of surface from AX, viz. \y. Hence 
the differential of the moment of BAD with reference to AX 







CENTRE OF GRAVITY. 



285 



is \y 2 dx, and the distance of the centre of gravity of BAD 

— fij 2 dx • 

from AX will be represented by . 

r J area 

Ex. 1. Determine the centre of gravity of 
an isosceles triangle. 

It is evident that the centre of gravity will A 
be somewhere in the line AD perpendicular 
to the base. 

Let AD=#, BD = 6, AR=a>, FU^y. 

Then yJ° 




a 



Hence 



AG: 



fxydx 



fix* 
J a 



dx 



2x 



area \xy 3 ' 

that is, the distance of the centre of gravity from the vertex 
of the triangle is equal to two thirds the altitude of the tri- 
angle. 

Ex. 2. Determine the centre of gravity of a parabolic area. 



Here 



2xy 4 i 3 
y'^&aXj and area = -~- = o& # ; 



AG: 



4 1 5 

13 — I "" 

fxydx f2a 2 x 2 dx 5 



a 2 x^ 



area 



area 



4 1 3. 
oO^X' 2 



3x 

: ~5> 



that is, the distance of the centre of gravity from the vertex 
equals three fifths of the axis of the parabola. 

Ex. 3. Determine the centre of gravity of a seg- 
ment of a circle. 

We will determine the distance of the centre of A ' 
gravity from the centre of the circle, instead of 
from A. 




The equation is 
Hence OG = 



yl _ T 2 _ 



fx(r 2 -x 2 ) 2 dx ^{r 2 -x 2 f+G 



area 



area 



286 



INTEGKAL CALCULUS. 



If the segment is a semicircle, then, integrating between 
x=0 and x=—r, and placing area=-f7ir 2 , we have 

Ex.4. Determine the centre of gravity of a segment of an 
ellipse cut off by an ordinate to the transverse axis. 



The equation is 



r=^0 2 -* 2 )- 



" 2 W 

a? 



Hence OG = 



f-(a?-xrfxdx —(a 2 -x 2 y+G 



OG= 



3tt' 





area area 

If the segment is a semi-ellipse, then, integrating between 
x=0 and x~ — a, and placing area=^7r^,we have 

4# 



Ex. 5. ABD is a segment of a parabola cut 
off by an ordinate to the axis. Determine 
the distance of its centre of gravity from AX. 

Ans. -*. 

Ex. 6. Determine the distance of the centre of gravity of 
ABE from AY. 3x 

Ans. Tq . 

Ex. 7. Determine the distance of the centre of gravity of 
ABE from AX. 3y 

4 

240. To find the centre of gravity of a surface of revo- 
lution. 

The differential of the surface generated by 
the revolution of AB around AX is 2iryds, 
and the differential of its moment in reference 
to the plane passing through AY perpendic- 
ular to AX is %nxyds. Hence the moment 
of the entire surface generated will be repre- 
sented by f2wxyd8, and the distance of the 



CENTRE OF GRAVITY. 287 

centre of gravity from AY will be represented by 

fiirxyds 

surface ' 

Ex. 1. Determine the centre of gravity of the convex surface 

of a right cylinder. 

Here y =b, and the convex surface of the cylinder =27rbx. 

_ T . _, f2wbxdx x 
Hence AGr— * 7 = -; 

2ttox 2 ' 

that is, the centre of gravity is at the middle point of the 
axis. 

Ex. 2. Determine the centre of gravity of the convex surface 
of a right cone. 

Here y=a#, and the convex surface of the cone=7ry(x 2 +y 2 )*. 

Also, ds = (dx 2 + dy 2 f = dx(a 2 + If. . 

A ^ f27rxy(a 2 + l)^dx 2x 
Hence AG=- yv \ =-»; 

irax 2 (a 2 + l) 2 ' 6 
that is, the centre of gravity is at a distance from the vertex 
equal to two thirds the length of the axis. 

Ex. 3. Determine the centre of gravity of the surface of a 
spherical segment. 

Here y 2 = r 2 —x 2 , 

Also, ds=(dx 2 +dy 2 ) 2 = y ~^~T 2] ^x- 

and surface — 2irrx. 

, „ [2-KTxdx x 
Hence AG=—^ =- 

2irrx 2 

T 

For a hemisphere, AG— ~; that is, the centre of gravity of 

the convex surface of a hemisphere is at a distance from the 
centre equal to half the radius. 

241. To find the centre of gravity of a solid of revolution. 
The differential of the solid generated by the revolution of 



1 

k 2 rdx 

'IT* 



288 



INTEGRAL CALCULUS. 



AB around AX is Try 2 dx, and the differen- 
tial of its moment in reference to a plane 
passing through AT perpendicular to the 
plane BAC is irxy 2 dx. Hence the moment 
of the solid generated will be represented 
by f7rxy 2 dx, and the distance of the centre 
of gravity from AY will be represented by 

fnxy 2 dx 

volume * 

Ex. 1. Determine the centre of gravity of a right cone. 

TT _ . iry 2 x 

Here y=ax, and volume z=—^-~. 

fira 2 x^dx 3x 




Hence 



AG: 



-gird x 



that is, the distance of the centre of gravity from the vertex is 
equal to three fourths of the axis. 

Ex. 2. Determine the centre of gravity of a paraboloid. 

Here y 2 —^ax^ and volume— \-Ky 2 x. 

f^irax 2 dx 2x 



Hence 



AG=^ 



2irax 2 ~ 3 ' 

that is, the distance of the centre of gravity from the vertex is 
equal to two thirds of the axis. 

Ex. 3. Determine the centre of gravity of a segment of a 
sphere. 



Here 
Hence 



y 2 = 2rx—x 2 , and volume: 



AG: 



fnx(2rx — x 2 )dx 



-irTX 1 

x(8r- 



-inx 3 . 
-Sx) 



wTX 2__^ 7rX 3 



4(3?'— #)' 



When x=r, 



AG=4; 



that is, the centre of gravity of a hemisphere is at a distance 
from the centre equal to three eighths of the radius. 

Ex. 4. Determine the centre of gravity of a segment of a 
prolate spheroid. 

b 2 
Here y 2 ~ - 2 (2ax— x 2 ). 

CO 




CENTRE OF GRAVITY. 289 

/ 7 ^Yx(2ax — x 2 )dx — { %ax*—\x* \ 
CO CO 

ilence Air = -, = ; — . 

volume volume 

When x = a, volume = 3 , 

and .AG=-g-; 

that is, the centre of gravity of a hemispheroid is the same as 
that of the hemisphere whose diameter is the major axis of 
the spheroid. 

Ex. 5. ABD is a segment of a parabola cut 
off by an ordinate to the axis. If this seg- 
ment revolve about BD, determine the dis- 
tance of the centre of gravity of the solid 

from AX. 5y 

Ans. lg . 

Ex. 6. If the segment ABE revolve about AY, determine 

the distance of the centre of gravity of the solid from AX. 

5y 
Aiis. -jr. 
o 

Ex. 7. If the segment ABE revolve about BE, determine 
the distance of the centre of gravity of the solid from AY. 

Ans. k. 
5 

242. General theorem. The examples in the six preceding 
articles will be recognized as special cases of a general theo- 
rem, which (using three co-ordinate axes) may be thus stated : 
The distance of the centre of gravity of a mass {in) from the 
plane yz is fxdm 

/dm** 
where dm is the differential of mass, and the two integrals 
have the same limits depending upon the extent of m. The 
distances of the same point from the other two co-ordinate 

planes are fydm . fzdm 

-y-j — and '—rj — . 
jam Jam 

N 



290 INTEGRAL CALCULUS. 



CENTRE OF OSCILLATION OF A PENDULUM. 

243. A simple pendulum. A single material particle sus- 
pended by a slender wire supposed to have no weight, and 
oscillating freely about a fixed horizontal axis connected with 
the other extremity of the wire, is called a simple pendulum. 

A compound pendulum. Any material body free to oscillate 
about a fixed horizontal axis is called a compound pendulum. 

A simple pendulum, as above defined, can have no physical 
existence; but it is convenient to regard a compound pendu- 
lum as a collection of simple pendulums. 

The axis of a compound pendulum is the straight line drawn 
through its centre of gravity perpendicular to the horizontal 
axis about which the pendulum vibrates. 

244. The centre of oscillation of a pendulum is that point 
of its axis, or of its axis produced, at which, if the entire mass 
of the pendulum were collected, its time of vibration would be 
unchanged. 

The length of a compound pendulum is that part of its axis 
which is included between the axis of suspension and the centre 
of oscillation, or it is the length of the simple pendulum which 
vibrates in the same time. 

245. Principle of moments. The resistance to the motion 
of a solid body about any axis is called the moment of 
inertia of the body with respect to that axis, and is ex- 
pressed by the sum of the products of each of the parti- 
cles of the body into the square of their respective dis- 
tances from the axis of rotation. Thus, let S be the point 
of suspension of a pendulum ; let A, B, C, etc., denote the 

component particles of the body, and a, &, c, etc., their re- 
spective distances from the axis of suspension. Then 

Aa 2 + B£ 2 + Cc 2 + etc. 
is called the moment of inertia of the body with respect 
to the point S. 






B 



CENTRE OF OSCILLATION OF A PENDULUM. 291 

246. The moment of the mass with respect to the axis of 
suspension is the product arising from multiplying the sum of 
the particles by the distance of the common centre of gravity 
from the axis of suspension. If G denote the centre of grav- 
ity of the pendulum, and SG the perpendicular from G upon 
the axis, then (A + B + C+ . . . ) x SG 

denotes the moment of the mass with respect to the axis. 

247. General formula. It is proved in treatises on Mechan- 
ics that the distance from the centre of oscillation to the axis 
of suspension is found by dividing the moment of inertia with 
respect to that axis by the moment of the mass with respect to 
the same axis. If, then, O represent the centre of oscillation 
of a pendulum SGD, and SO the distance of O from the axis 
of suspension, we shall have 

Aa 2 + B6 2 + Cc 2 + etc. 



SO = 



(A+B + C+...)xSG- 



248. If we represent one of the elementary portions of the 
body, that is, the differential of the body, by dm, and its dis- 
tance from the axis of suspension by x, then x 2 dm will repre- 
sent its moment of inertia, and/afti^ will denote the moment 
of inertia of all the particles of the body. If we represent 
the mass of the body by ra, and the distance of the centre of 
gravity from S by g, we shall have 

qn __ fx 2 dm fx 2 dm 
~~ fxdm ~~ mg 

Ex. 1. Find the centre of oscillation of a slender rod or 
straight line suspended from one of its extremities. 

Here, as we regard only the length of the rod, dm will be 
represented by dx. Hence 

so, ^ . x3 

mg 3?ng' 
If x be taken equal to the whole length of the rod, then 

m~ x, and ^==-5. 



292 



INTEGRAL CALCULUS. 



Hence 



S0 = 



2x 



• s 



that is, the distance of the centre of oscillation from the axis 
of suspension is two thirds the length of the rod. 

Ex. 2. If a cylindrical rod 58.65 inches long, when suspend- 
ed from one extremity, make a vibration in one second, what 
is the length of the seconds' pendulum ? Ans. 39.1 inches. 
Ex. 3. Find the centre of oscillation of a slender rod sus- 
pended from any point above its centre of gravity. 

Let S be the point of suspension of the rod AP. The 
moment of inertia of the particles in SA is fx 2 dx=^SA 3 ; 
and the moment of inertia of the particles in SP is -3-SP 3 , 
the sum of which is ^(SP 3 + SA 3 ). The vibrating body is 
o SP + SA, and the distance of the centre of gravity from 
S=£(SP-SA). Therefore 

m^(SP 2 -SA 2 ). 
^SP 3 + SA 3 ) 2 SP» + SA3 
nence &u -^ S p 2 _ S A 2 ) ~ 3 x SP 2 - SA 2 ' 

Ex. 4. If a cylindrical rod 64 inches long, whose axis of sus- 
pension is distant 6 inches from one ' extremity of the rod, 
make a vibration in one second, what is the length of the 
seconds' pendulum? Ans. 39.13 inches. 

249. It is proved in treatises on Mechanics that along a 
given line through the centre of gravity 
the distance of the centre of oscillation of 
a pendulum from the centre of gravity 
varies inversely as the distance of the 
point of suspension from the same. If 
ABC be a solid of revolution, G its centre 
of gravity, O the centre of oscillation when 
the body is suspended from A, and O' the 
centre of oscillation when it is suspended 
from S, then 

GO:GO'::SG:AG. 
Hence, if we can find the centre of oscillation of the body 




CENTRE OF OSCILLATION OF A SOLID OF REVOLUTION. 293 

when suspended from any point A, we can find the centre of 
oscillation when it is suspended from any other point situated 
on the line AG. 



250. Centre of oscillation of a solid of revolution. Sup- 
pose a solid to be generated by the revolution of the curve 
BAC about its axis AM ; it is required to find its centre of 
oscillation when suspended from its vertex A. 

Let ENRF be a circular section 
perpendicular to the axis AM; let 
EF be that diameter of the section 
which is parallel to the axis of sus- 
pension ; draw NR parallel, and MP 
perpendicular to EF. Join AP, MR. 
We w T ish to find the moment of in- 
ertia of the circle ENRF regarded as a plate of matter whose 
thickness is indefinitely small. 

Let £M=a, NR=r, MF=v. 

Then AP 2 = a 2 + v 2 , NR = 2 V^v 2 , 

and the differential of the area ENPRF is 

dA—2dvVr 2 —v 2 . 
Hence the moment of inertia of an elementary portion of the 
circle is 2dv Vr 2 — v 2 x (a 2 + v 2 ), 

or 2a 2 dv Vr 2 —v 2 + 2v 2 dv Vr 2 —v 2 . 

Hence the moment of inertia of the circle ENRF is 




a 2 A+/2v 2 dvVr' 

7' 2 A V 

a 2 A+-^-^(r 2 -v 2 ) 
When v=r, this expression becomes 



or 



3 



V* 2 + x) x semicircle ENRF, 



and for the whole circle the expression becomes 



(« 2 +9 7r?,2= 



ira 2 r 2 + \wr\ 



Let AM=«, EM or MB=y. Then, for the circular section 



294 



INTEGKAL CALCULUS. 



EKRF the moment of inertia is w% 2 y 2 +&•}/*, and therefore 
the moment of inertia of the entire solid with respect to its 
vertex A will be f(Try 2 x 2 dx+^Try*dx), (1) 

where the integration extends to the whole solid. 

In order to apply this formula to a particular case, we must 
substitute for y its value in terms of %, as determined from 
the equation of the curve ; the integral will give the moment 
of inertia of the solid. 

Ex. 1. Find the centre of oscillation of 
a sphere suspended from its vertex A. 

Let r=the radius of the sphere, and let 
O be the centre of oscillation when it is 
suspended from A. 

The equation of the curve is 
y 2 = 2rx—x 2 . 
Hence the moment of inertia equals 
f7rx 2 dx(2i"x — x 2 ) + f\irdx{2rx — a? 2 ) 2 , 




which equals 



7T? )2 X 3 TTTX* 3ttX 5 



Let x=2r. 
have 

Hence 



3 V-4 20 " 

Then, since m=^7rr 3 , and <?=?', by Art. 248 we 

§7TT 5 + 4:7TT 5 - 
AO~ T -T 



^irr* 



GO=%r. 



Ex. 2. Find the centre of oscillation of a sphere suspended 
from any point S. 

Let O' be the centre of oscillation when suspended from S, 

2r 
Then, by Art. 249, T : GO' : : SG : r. 



Hence 



2 T 2 
G0 ' = 5*SG- 



Ex. 3. A sphere two inches in diameter, suspended by a slen- 
der wire whose length, measured from the point of suspension 
to the centre of the sphere, was 39.1 inches, made one vibration 
per second. What is the length of the simple seconds' pen- 
dulum ? Ans. 39.1102 inches. 



ON THE ATTRACTION OF BODIES. 295 



ON THE ATTRACTION OF BODIES. 

251. Units of measurement. In order to compare forces 
with each other, it is necessary to employ some standard units 
of time, space, velocity, etc. One second is usually taken for 
the unit of time, and one foot for the unit of space. In treat- 
ing of the heavenly bodies, it is sometimes, however, more con- 
venient to employ other units of time and space. 

Uniform velocity is measured by the number of units of 
space described in a unit of time ; that is, by the number of 
feet described in one second. When the velocity is variable, 
it is measured by the limit of the ratio of the space described 
to the time of describing it. 

The English unit of mass is the standard pound (avoirdu- 
pois). The mass of any given substance is denoted by the 
number of pounds it contains. 

252. Momentum. The momentum of a moving body, or 
its quantity of motion, is the product obtained by multiplying 
the mass moved by the velocity with which it is moved, and is 
expressed numerically by the product of the number of units 
of mass which it contains, and the number of units of velocity 
in its motion. Thus, if a body contain five pounds of matter, 
and move at the rate of eight feet per second, its momentum 
is represented by the number forty. 

253. Measure of force. Our simplest conceptions of force 
are derived from weight. The force with which the earth 
attracts the unit of mass at a given place is the usual unit of 
force. But in Mechanics other units of force are often more 
convenient. A constant force is measured by the momentum 
it communicates to a body in a unit of time. A variable force 
is measured at any instant by the momentum it would com- 
municate if it continued constant for a unit of time from the 
given instant. The unit of force is that force which, if con- 
tinued constant, would impart in a second of time a velocity 
of one foot per second to a pound of matter. 



296 



INTEGRAL CALCULUS. 



If the mass moved contain m pounds of matter, and if the 
velocity generated in a unit of time is represented bjf the 
units of force will be represented by mf 

254. Law of gravitation. According to the Newtonian law 
of gravitation, two particles of matter placed at any sensible 
distance apart, attract each other with a force which varies 
directly as the product of their masses, and inversely as the 
square of their distance from each other. 

Suppose, then, a particle to be attracted by all the particles 
of a body; if we resolve the attraction of each particle into 
components parallel to fixed rectangular axes, and take the sum 
of the components which act in a given direction, we shall ob- 
tain the resolved attraction of the whole body on the particle 
in that direction, and can thus ascertain the resultant attraction 
of the body in magnitude and direction. We shall apply this 
principle to a few simple cases. 

255. To find the resolved attraction of a uniform straight 
line which is exerted in a direction perpendicular to the 
line. 

By a straight line we understand a material prism or cylin- 
der whose section perpendicular to its axis is indefinitely small. 
Let AB be the given straight line, and 
P a particle of matter attracted by AB. 
Draw PA perpendicular to AB ; let YK=a ; 
represent AC, a variable part of AB, by ,r. 
Then TC=Va 2 +x\ 

Let m denote the mass of a portion of the given line one 
unit in length ; then mdx will denote the mass of an element- 
ary portion of the line AB. Let m! denote the mass of P ; 
then, by Art. 254, the attraction of the elementary portion at 
C upon the particle P will be 

Icmm'dx 
PC 2 ' 
where k denotes a constant quantity, and is equal to the attrac- 
tion of a unit of mass upon a unit of mass placed at a distance 




ON THE ATTRACTION OF BODIES. 297 

equal to the unit of length. This attraction is exerted in the 
direction of the line PC. By the principle of the resolution 
of forces, the force of attraction toward C : the force toward 
A : : PC : PA : : Va? + x 2 : a. Therefore the resolved part of the 
attraction of an elementary portion of the line AB at C ex- 
erted in the direction PA is 

Tcmm'adx 

which is therefore the differential of the attraction of AC for P. 
The integral of this expression is 

hnm'x „ 
aVtr + x 2 
which denotes, therefore, the whole attraction of AC upon P 
exerted in the direction AP. 

To determine the value of the constant C, make #=0; the 
attraction of AC will then = 0. Hence C = 0. 

If we make #=AB, the above expression becomes 
hm/m! x AB 
PAxPB ' 
which denotes the attraction of the whole line AB for P ex- 
erted in the direction AP. 

256. To find the attraction of a uniform circular lamina 
on a particle situated in a straight line draivn through the 
centre of the circle perpendicular to its plane. 

Suppose O to be the centre of 
the circle ABCD, and a particle of 
matter to be situated at P in the 
line PO drawn through O perpen- 
dicular to the plane of the circle. 
Take EO, any part of AO, and let 
another circle be described with O 
as a centre, and EO as the radius. 

Let PO = a, a nd PE = x . 

Then EO=Va^a" 2 . 

N2 




298 INTEGRAL CALCULUS. 

The area of the circle whose radius is EO is tt(x 2 — a 2 ), and 
its differential is 2-n-xdx. If we put m to denote the mass 
of a square unit of the lamina, and ml the mass of P, then, 

as in Art. 255, p™ will denote the attraction of the 

particles of the elemental ring for P. But the force of attrac- 
tion of any particle : the force toward O : : PE : PO. Hence 
the differential of the attraction of the circle for P exerted in 
the direction PO is 

2kmm f 7rxdx a 2kmm f iradx 

tAs tAs tAs 

The integral of this expression is 

2hmm r Tra 



fC 



x 

When x~a^ EO = 0, and the attraction of the circle whose 
radius is EO^O. Hence 

= — 2Jcmm , 7r + C ; that is, C = 2Jcmm , 7r. 
Hence the attraction of the circle for P is 

2hnm f 7r(l — -). 
When #=PA, the preceding expression becomes 

2fanm / 7r[l — pxjj 

which represents the attraction of the whole circle ABCD on 
the particle P. For greater convenience, we will represent 
the constant factor 2hmm'iT by KA, where A represents the 
thickness of the lamina. Hence the attraction of the circle 

ABCD on the particle P is KAU-jrg-Y 

If we suppose the radius of the circle to become infinite, we 
obtain for the attraction of an infinite lamina on an external 
particle the constant quantity KA; that is, the attraction is in- 
dependent of the distance of the attracted particle from the 
lamina, and is directly as the thickness of the lamina. 



ON THE ATTRACTION OF BODIES. 



299 



257. To find the attraction of a homogeneous sphere upon 
a particle situated without it. 

Let O be the centre of the 
sphere, and P the attracted 
particle. Through P and O 
draw the line POB, and let 
CED be a section of the 
sphere perpendicular to AB. 
Let AO = a, PO = 6, PA=<?=5-0, PE=y, VC = c+x. 




Then 
Now 

that is, 

Whence 



AE=y—c, and EB = 2a — y + c. 
AExEB = EC 2 = PC*-PE 2 ; 



(y-c)(2a-y + c) = (c + x) 2 - 
2ac + 2c 2 + 2cx + x 2 



-y- 



y- 



2a + 2c 

2bc + 2cx + x 2 



2b 



, since b = a + c. 



Hence, by Art. 256, the attraction of the circle CED upon P is 



equal to K A 



V o+xP 



or 



or 



c+x/ 



2bc + 2cx+x 2: 



2b(c+x) 
I2ax—x 2 



jfy, 



(c+x). 



\ 
)dy. 



Hence the differential of the attraction of the sphere is equal to 

'K{2ax—x 2 )dy 

cdx + xdx 



or, since dy=.- 
sphere is 

whose integral is 



2b{c + x) ' 
-, the differential of the attraction of the 



K(2ax—x 2 )dx 



Kx 



2b 2 

ax 2 — U 



2b 2 



C. 



When a? = 0, the attraction becomes =0, and therefore C = 0. 
Hence the entire expression for the attraction of the segment 



ACD upon P is 



Kx 



ax* 



■\# 



2* 2 



300 



INTEGRAL CALCULUS. 



Let x=2a, and we shall have the attraction of the whole 



sphere: 



2a 3 K 



o A2~> which varies as -p. Now the contents of spheres 

vary as a 3 ; hence the attraction of a sphere varies as the quan- 
tity of matter divided by the square of the distance from the 
centre. The attraction of a sphere for a particle of matter 
situated without it is therefore the same as if all the matter 
of the sphere were collected into its centre. 

258. Velocity of a body when the motion is uniform. When 
the motion of a body is uniform, the space described in any 
time is equal to the product of the numbers which express the 
velocity and the time. 

Let v denote the velocity expressed in feet, t the time ex- 
pressed in seconds, and s the space described. Then s = tv; 

whence v—-^ or the velocity is the quotient of the space di- 
vided by the time. 

259. Velocity of a body when the motion is variable. When 
the velocity is not constant, it can no longer be measured by 
the quotient of the space divided by the time ; for these quo- 
tients are different for different times. 

Let the time of motion be represented by the line BF, and 
let us suppose it to be divided into any 
number of equal portions BC, CD, DE, 
etc. Draw the ordinates BJ, Cc, Dd, etc., 
which shall represent the velocities of 
the body at the instants corresponding 
to the points B, C, D, etc. ; and complete 
the parallelograms bC, cD, g?E, etc. If 
the body were to move daring the time BC with the velocity 
which it had at the beginning of that interval, the space de- 
scribed would be proportional to the rectangle bC. Also, if 
during the time CD the body were to move with the velocity 
which it had at the beginning of that interval, ths epace de- 







ON THE ATTRACTION OF BODIES. 301 

scribed would be proportional to the rectangle cD ; and sim- 
ilarly for the other portions. 

Therefore, if the body were to move during each portion of 
time with the velocity which it had at the beginning of that 
portion, the whole space described would be proportional to 
the sum of the rectangles £0, dD, 6?E, etc. 

If the number of these portions of time be increased indef- 
initely, and consequently the magnitude of each portion be 
diminished indefinitely, the space thus described will have for 
its limit the space described in the whole time with the varia- 
ble velocity. But the sum of the rectangles JC, cD, etc., has 
for its limit the curvilinear area &BF/*; hence the space de- 
scribed in the whole time will be proportional to the curvilinear 
area bBFf. 

Complete the parallelogram 5BFM. Then, since 
6BxBF = £BFM, 

m 5BFM bBFf JBFM 
we have bB = -^jr-= --gjr- x j^j. 

6BFM . 
Now, when BF vanishes, the limit of zppy? is unity. Hence 

6B = the limit of ~rr&- when BF vanishes. 

But bBFf denotes the increment of the space, and BF the 

increment of the time. Hence 

, ... i t . r ^ ncr - °f space 

the velocity— the limit or t-t. : 

J mcr. oi time ' 

ds 
that is, v ~Tf ( ) 

260. To find the accelerating force ivhen it is constant. 
When the accelerating force is uniform, the velocity generated 
in any time is equal to the product of the numbers represent- 
ing the force and time. Let/ 1 be the accelerating force, or the 
velocity generated in one second. Since the force is uniform, 
f will be the velocity added in the next second, and 2/* will be 
the velocity at the end of two seconds; Sf will be the velocity 



302 INTEGRAL CALCULUS. 

at the end of three seconds, and tf will be the velocity at the 

v 
end of t seconds. Hence v — tf or f=z-; that is, the accelerat- 

t 

ing force is equal to the velocity divided by the time. 

261. To find the accelerating force when it is variable. 

When the accelerating force is not constant, it may be shown 

(as in Art. 259) that it is measured by the limit of the quotient 

of the increment of the velocity divided by the corresponding 

increment of the time, when these increments ultimately vanish ; 

... . . n _. . _ incr. of velocity 

that is, the accelerating torce = limit of —. j— -. , 

° , mcr. or time ' 

/=$ » 

If we differentiate Eq. (1), Art. 259, with respect to v, we shall 
have d 2 s 

dv =dt' 

Hence f = ~dt 2 ' ^ 

By combining equations (1) and (2), we find 

fds — vdr. (4) 

Equations (1), (2), (3), (4) are called the differential equations 
for force and motion, and may be applied to the solution of 
a great variety of problems. 

262. Falling bodies near the earth! s surface. At any given 
place, and within the range of small elevations, gravity may be 
considered as a constant accelerating force. Substituting g 
for f in Eq. (2), we have dv—gdt. 

Integrating, we have v — gt + C. 

. If the motion of the body commence with the time, then, 
when t=0, v = y and consequently C = 0. Hence 

v=gt, and t=~. (5) 

Substituting the value of v in Eq. (1), we have 

ds=vdt=gtdt. 






ON THE ATTRACTION OF BODIES. 303 

at 2 
Integrating, we have s = -5- + C. (6) 

If the motion of the body commence with the time, C = 0, and 

we have .=£. 




Whence t = \ —. (7) 



From Equations (5), (6), and (7), when either the space, time, 
or velocity of a falling body is given, the other two quantities 
can be computed. 

263. If we wish to find the space described during any speci- 
fied portion of the time of fall, it is only necessary to take the 
integral in Eq. (6) between the required limits. If, for ex- 
ample, we wish to find the space described by a falling body 
in the 20th second from the beginning of the fall, we have 

S =f( 2 0>-19*>= 3 -|2. 

If a body be projected vertically downward with a velocity 
v\ the space described in the time t will be 

, gt 2 

s=vt+-^-. 

If the body be projected vertically upward with a velocity v', 
the space described in the time t will be 

264. Fall of bodies through great distances. According to 
Art. 257, the earth's attraction (that is, the force of gravity) 
varies inversely as the square of the distance from the centre 
of the earth. Let r—the radius of the earth, g the force of 
attraction at its surface, a the distance of the body from the 
centre at the commencement of its motion, and x its distance 
at the end of t seconds. 



304 INTEGEAL CALCULUS. 

Then f-9^'%- 

Whence fJ^. (8) 

265. To find the velocity acquired. 
ByEq.(4), fds=vdv. 

Whence /=*£=£> (frm Eq. (8)). 

But since s=a—x, ds= —dx. 

Hence vdv = — 9 . 

ar 

2^r 2 
Integrating, we have ?; 2 = — — + C. 

x 

The constant C depends upon the velocity which the body 
has at the distance a where gravity begins to act. If this ve- 
locity is zero, then 2gr 2 

a 

Whence C=-^. 

a 

Hence the velocity acquired in falling from the initial distance 

a to the distance x is given by the expression 

2 2gr 2 ^ 2gr 2 2gr\a—x) 

X CO CLX 

When the body arrives at the surface of the earth (that is, 
when x—t\ its velocity will be 



v=y 



2gr{a — r) 



a 

Ex. 1. Find the velocity acquired by a body falling to the 
earth from the distance of the moon. 

Assuming the mean radius of the earth to be 3956 miles, 
and the distance of the moon equal to 60 radii of the earth, 
we shall have a=^Q0r. Hence 



<w 



59gr 



30 - 
The value of g is 32i feet. Hence v—6.88 miles per second. 



ON THE ATTRACTION OF BODIES. 305 

Ex. 2. Find the velocity acquired by a body falling to the 
earth from the height of 100 miles. 

Am. 5755.9 feet, or 1.09 mile per second. 

If a be infinite, then =1, and we have 

a 

v=V2gr=Q.94: miles per second. 
Therefore the greatest possible velocity acquired by a body in 
falling to the earth is less than seven miles per second, and a 
body projected upward with that velocity (supposing there is 
no resisting medium) would never return. 

266. To find the ti?ne of falling. 
ByEq.(l),Art.259, dt~. 

Substituting a— x for s, and expression (9) for v, we obtain 

dt— — x d(a—x) 

V2gr 2 (a—x) 

<Ja xdx 

— x — 



rV2g Vax—x 2 

___. -\/a r —xdx 

Whence t = — 7= / — . 

rV2g J Vax—x 2 

Integrating by Formula E, Art. 196, we have 



J \j 



— xdx y = a . _2x „ 

v ax— x 2 — 7. ver. sin. ] — + U. 



Vax—x 2 * a 

li t=0 when x=a, we have 7ra 

__ y/a ( / a • -i 2 ^ 7ra ) 

Hence t— — z=< Vax— x 2 — 5 ver. sin. ■ — + -^- K 
rV2g I 2 a 2 ) 

which denotes the number of seconds elapsed in moving from 

the distance a to the distance x from the centre of the earth. 

When x—r (that is, when the body arrives at the surface of 

the earth), the number of seconds elapsed will be 

V® ( / 5 a • -t 2 ?' ira ) 

t— — 7= \ Yar—r 2 —n ver. sin. 3 — +-~- > . 
rV2g \ * a a ) 



306 INTEGRAL CALCULUS. 

The last two terms can be united into a single term. Since 
7T— arc whose ver. sin. is A = arc whose cos. is (A— 1), 

. ,2r J2r \ <2r-a\ 

7T— ver. sm." 1 — =cos. I — — 1 J^cos. -1 ! J. 

a \a I \ a / 

Hence t— — ;= \ Var—r 2 + 7: cos. _1 ( ] 

rV2g I 2 \ a ) 

Ex. 1. Find the time required for a body to fall to the earth 
from the distance of the moon. 

Assuming the same values of &, g, and r as in Art. 265, we 
find £ = 415,600 seconds, which equal 4 days, 19 hours, 26 
minutes, and 40 seconds, which is the time a body would be 
in falling from the moon to.the earth's surface. 

Ex. 2. Find the time required for a body to fall to the earth 
from a height equal to the radius of the earth. 

Ans. 2072 seconds = 34 min. 32 sec. 

Ex. 3. Find the time required for a body to fall to the earth 
from the height of 100 miles. Ans. 185 seconds. 

If a be infinite, then t is infinite, although the velocity, as 
found in Art. 265, is less than seven miles per second. 

267. To find the attraction of a homogeneous spherical shell 
of small thickness on a particle situated within it. Let ABCD 
be a spherical shell, and P a particle situ- 
ated at any point within it. Suppose P to 
be the vertex of a double cone, whose angle 
is indefinitely small, and let the cone cut 
the shell in AB, ab. The inclination of the 
bases of the cone to the axis of the cone 
will be the same. Hence the bases of the 
cones are similar figures ; and since their thickness is the same, 
the quantity of matter in the bases will vary as AB 2 : ab 2 . Let 
Q and q represent the quantity of matter in the bases. Then 
Q: ? ::AB 2 :^ 2 ::AP 2 :6P 2 ; 
Q q 

that is, AP 2 ~~ bP 2 ' 




ON THE ATTRACTION OF BODIES. 307 

Q Q 

But xpa an( i rpi are proportional to the attractions of the 

bases on the particle P; and since these, attractions are equal 
and in opposite directions, they will not affect the particle P. 
The same will evidently apply to two spherical polygons ba 
and AB, the bases of any two opposite spherical pyramids, and 
the whole shell may thus be divided into p^iirs of elements at- 
tracting equally and in opposite directions; and therefore a 
particle within the shell is equally attracted in every direction, 
and hence this attraction can have no effect in drawing the 
particle in any one direction more than in another. 

268. Inside a homogeneous sphere the attraction varies as 
the distance from the centre. Suppose a particle situated in- 
side a homogeneous sphere, at the distance x from its centre. 
Then, by Art. 267, all that portion of the sphere which is at a 
greater distance than x from the centre produces no effect upon 
the particle ; also, the remainder of the sphere attracts the par- 
ticle in the same manner as if the mass of the remainder were 
all collected at the centre of the sphere. Let. Q represent the 
quantity of matter in this sphere. Then the attraction on the 

Q 

particle P is proportional to — ^ But Q is proportional to x 3 . 

X r> 

Hence the attraction on the particle varies as — , or x ; that is, 

the attraction varies as the distance from the centre of the 
sphere. 

269. To determine the velocity acquired by a body falling 
beneath the surface of the earth. Suppose an opening to be 
made from the surface to the centre of the earth, and that a 
body moves along it under the influence of the earth's attrac- 
tion. By Art. 268, the accelerating force f varies directly as 
the distance from the centre. Let r denote the radius of the 
earth, g the force of gravity at the surface, and s any variable 
distance from the surface. Then r—s will be the distance 
from the centre, and we shall have 



308 INTEGRAL CALCULUS. 

f:g::r-s:r, 

or f=^ -. 

But by Eq. (3), Art. 261, 

d 2 s g(r—s) 
J~dt % - r ■ 
Multiplying by 2ds, we obtain 

2dsd 2 s 2q, v , • 

Integrating, we obtain 

To determine the constant C, suppose the body to start from 
the surface without any initial velocity. Then, when s = 0, v = 0. 
Hence = — gr + 0, 

or Q—gr. 

Therefore v 2 = g -{r 2 -(r-sf}, (1) 



r K 



or 



^=\/!{^-(^-*) 2 }, 



from which we may deduce the value of v for any point in 
the line of the body's motion. 

If we make s=r, this equation becomes 

v—Vgr^ 
which is the velocity acquired by the body when it has arrived 
at the centre of the earth. From this point the body, in con- 
sequence of its acquired velocity, will ascend to the surface, 
and, having moved through an entire diameter, we shall have 
s =2r, which, substituted in the above equation, gives ^ = 0; 
that is, the velocity will be 0. It will therefore again return 
through the centre, and so on, making an infinite number of 
oscillations, each of equal duration. 

Ex.1. Compute the velocity acquired by a body in falling 
from the surface to the centre of the earth. 

Ans. 4.91 miles per second. 



ON THE ATTRACTION OF BODIES. 309 

Ex. 2. Compute the velocity acquired by a body in falling 
from the surface of the earth to the depth of 100 miles. 

Ans. 1.0968 mile per second. 

270. To determine the time of a falling body beneath the 
surface of the earth. 

From Eq. (1), Art. 269, we have 

'9 , 4 _ ds 



Hence \J y -.dt 



Integrating, we obtain 



y t 1 — (r — s)' 

It C ds 
t=\/ -x I , ■ , 



-(r—s) : 



r r—s 
xcos. -1 + U. 

g r 



When 5=0, t = 0, and hence C = 0. 

7r [r 
When s=r,we have t— ^V ~ 5 

2 v g' 

which denotes the time of falling from the surface of the 
earth to the centre. The time of falling through the diam- 
eter will be /- 

Ex. 1. Compute the time required for a body to fall from 
the surface to the centre of the earth. Ans. 21 min. 6 sec. 

Ex. 2. Compute the time required for a body to fall from 
the surface of the earth to the depth of 100 miles. 

Ans. 181.6 seconds. 



THE END. 



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